solubility and titration experiment

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Titrating sodium hydroxide with hydrochloric acid

In association with Nuffield Foundation

In this experiment students neutralise sodium hydroxide with hydrochloric acid to produce the soluble salt sodium chloride in solution. They then concentrate the solution and allow it to crystallise to produce sodium chloride crystals

You have to decide if this experiment is suitable to use with different classes, and look at the need for preliminary training in using techniques involved in titration (see Teaching notes). What follows here assumes that teachers have judged the class to be capable of doing this experiment using a burette with reasonable expectation of success.

Assuming that the students have been given training, the practical work should, if possible, start with the apparatus ready at each work place in the laboratory. This is to avoid vulnerable and expensive glassware (the burette) being collected from an overcrowded central location.

Source: © Getty Images

Students doing a titration experiment in a school science laboratory.

Time required

Filling the burette, measuring out the alkali into the flask, and titrating it until it is neutralised takes about 20 minutes, with false starts being likely for many groups. In practice it does not matter if the end-point is overshot, even by several cubic centimetres, but the aim is to find the proportions for a roughly neutral solution.

Producing a neutral solution free of indicator, should take no more than 10 minutes.

Evaporating the solution may take the rest of the lesson to the point at which the solution can be left to crystallise for the next lesson. Watching solutions evaporate can be tedious for students, and they may need another task to keep them occupied – eg rinsing and draining the burettes with purified water.

• Eye protection
• Burette, 30 or 50 cm 3 (note 1)
• Conical flask, 100 cm 3
• Beaker, 100 cm 3
• Pipette, 20 or 25 cm 3 , with pipette filter
• Stirring rod
• Small (filter) funnel, about 4 cm diameter
• Burette stand and clamp (note 2)
• White tile (optional; note 3)
• Bunsen burner
• Pipeclay triangle (note 4)
• Evaporating basin, at least 50 cm 3 capacity
• Crystallising dish (note 5)
• Microscope or hand lens suitable for examining crystals in the crystallising dish

Apparatus notes

• If your school still uses burettes with glass stopcocks, consult the CLEAPSS Laboratory Handbook, section 10.10.1, for their care and maintenance. This experiment will not be successful if the burettes used have stiff, blocked or leaky stopcocks. Modern burettes with PTFE stopcocks are much easier to use, require no greasing, and do not get blocked. Burettes with pinchcocks of any type are not recommended; while cheap, they also are prone to leakage, especially in the hands of student beginners.
• Burette stands and clamps are designed to prevent crushing of the burette by over-tightening, which may happen if standard jaw clamps are used.
• The optional white tile is to go under the titration flask, but white paper can be used instead.
• Ceramic gauzes can be used instead of pipeclay triangles, but the evaporation then takes longer.
• The evaporation and crystallisation stages may be incomplete in the lesson time. The crystallisation dishes need to be set aside for crystallisation to take place slowly. However, the dishes should not be allowed to dry out completely, as this spoils the quality of the crystals. With occasional checks, it should be possible to decide when to decant surplus solution from each dish to leave good crystals for the students to inspect in the following.
• Sodium hydroxide solution, 0.4 M (IRRITANT), about 100 cm 3 in a labelled and stoppered bottle
• Dilute hydrochloric acid, 0.4 M, about 100 cm 3 in a labelled and stoppered bottle
• Methyl orange indicator solution (or alternative) in small dropper bottle

Health, safety and technical notes

• Read our standard health and safety guidance .
• Wear eye protection throughout.
• Sodium hydroxide solution, NaOH(aq), (IRRITANT at concentration used) – see CLEAPSS Hazcard  HC091a and CLEAPSS Recipe Book RB085. The concentration of the solution does not need to be made up to a high degree of accuracy, but should be reasonably close to the same concentration as the dilute hydrochloric acid, and less than 0.5 M.
• Dilute hydrochloric acid, HCl(aq) – see CLEAPSS Hazcard  HC047a and CLEAPSS Recipe Book RB043. The concentration of the solution does not need to be made up to a high degree of accuracy, but should be reasonably close to the same concentration as the sodium hydroxide solution, and less than 0.5 M.
• Methyl orange indicator solution (the solid is TOXIC but not the solution) – see CLEAPSS Hazcard  HC032  and CLEAPSS Recipe Book RB000. 

Source: Royal Society of Chemistry

Apparatus for titrating sodium hydroxide with hydrochloric acid to produce sodium chloride.

• Using a small funnel, pour a few cubic centimetres of 0.4 M hydrochloric acid into the burette, with the tap open and a beaker under the open tap. Once the tip of the burette is full of solution, close the tap and add more solution up to the zero mark. (Do not reuse the acid in the beaker – this should be rinsed down the sink.)
• Use a pipette with pipette filler to transfer 25 (or 20) cm 3  of 0.4 M sodium hydroxide solution to the conical flask, and add two drops of methyl orange indicator. Swirl gently to mix. Place the flask on a white tile or piece of clean white paper under the burette tap.
• Add the hydrochloric acid to the sodium hydroxide solution in small volumes, swirling gently after each addition. Continue until the solution just turns from yellow-orange to red and record the reading on the burette at this point. This coloured solution should now be rinsed down the sink.
• Refill the burette to the zero mark. Carefully add the same volume of fresh hydrochloric acid as you used in stage 1, step 3, to another 25 (or 20) cm 3  of sodium hydroxide solution, to produce a neutral solution, but this time without any indicator.
• Pour this solution into an evaporating basin. Reduce the volume of the solution to about half by heating on a pipeclay triangle or ceramic gauze over a low to medium Bunsen burner flame. The solution spits near the end and you get fewer crystals. Do not boil dry. You may need to evaporate the solution in, say, 20 cm 3  portions to avoid overfilling the evaporating basin. Do not attempt to lift the hot basin off the tripod – allow to cool first, and then pour into a crystallising dish.
• Leave the concentrated solution to evaporate further in the crystallising dish. This should produce a white crystalline solid in one or two days.
• Examine the crystals under a microscope.

Looking for an alternative method?

Check out our  practical video on preparing a salt  for a safer method for evaporating the solution, along with technician notes, instructions and a risk assessment activity for learners.

Teaching notes

Titration using a burette, to measure volumes of solution accurately, requires careful and organised methods of working, manipulative skills allied to mental concentration, and attention to detail. All of these are of course desirable traits to be developed in students, but there has to be some degree of basic competence and reliability before using a burette with a class. The experiment is most likely to be suited to 14–16 year old students. This is discussed further below, but what follows here assumes that you have judged the class to be capable of doing this experiment using a burette with reasonable expectation of success.

Students need training in using burettes correctly, including how to clamp them securely and fill them safely. You should consider demonstrating burette technique, and give students the opportunity to practise this. In this experiment a pipette is not necessary, as the aim is to neutralise whatever volume of alkali is used, and that can be measured roughly using a measuring cylinder.

It is not the intention here to do quantitative measurements leading to calculations. The aim is to introduce students to the titration technique only to produce a neutral solution.

Alternative indicators you can use include screened methyl orange (green in alkali, violet in acid) and phenolphthalein (pink in alkali, colourless in acid).

Leaving the concentrated solutions to crystallise slowly should help to produce larger crystals. The solubility of sodium chloride does not change much with temperature, so simply cooling the solution is unlikely to form crystals.

Under the microscope (if possible, a stereomicroscope is best) you can see the cubic nature of the crystals. If crystallisation has occurred in shallow solution, with the crystals only partly submerged, ‘hopper-shaped’ crystals may be seen. In these crystals, each cube face becomes a hollow, stepped pyramid shape.

Student questions

What substances have been formed in this reaction? Write a word equation and a symbol equation.

Why must you use another 25 cm 3  of sodium hydroxide solution, rather than making your crystals from the solution in stage 1?

What shape are the crystals?

More resources

Inspire learners and discover more ways chemists are making a difference to our world with our video job profiles .

Additional information

This is a resource from the  Practical Chemistry project , developed by the Nuffield Foundation and the Royal Society of Chemistry.

Practical Chemistry activities accompany  Practical Physics  and  Practical Biology .

The experiment is also part of the Royal Society of Chemistry’s Continuing Professional Development course:  Chemistry for non-specialists .

© Nuffield Foundation and the Royal Society of Chemistry

• 14-16 years
• 16-18 years
• Practical experiments
• Practical skills and safety
• Acids and bases

Specification

• 1.8.18 demonstrate knowledge and understanding of how pure dry samples of soluble salts can be prepared by: adding excess insoluble substances to acid; adding alkali to acid, or vice versa, in the presence of an indicator; and repeating without indicator…
• 8. Investigate reactions between acids and bases; use indicators and the pH scale
• Mandatory eexperiment 4.2A - A hydrochloric acid/sodium hydroxide titration, and the use of this titration in making the sodium salt.
• 3. Find the concentration of a solution of hydrochloric acid
• 2a Determination of the reacting volumes of solutions of a strong acid and a strong alkali by titration.
• The volumes of acid and alkali solutions that react with each other can be measured by titration using a suitable indicator.
• Students should be able to: describe how to carry out titrations using strong acids and strong alkalis only (sulfuric, hydrochloric and nitric acids only) to find the reacting volumes accurately
• Salt solutions can be crystallised to produce solid salts.
• Students should be able to describe how to make pure, dry samples of named soluble salts from information provided.
• 5.9C Carry out an accurate acid-alkali titration, using burette, pipette and a suitable indicator
• 3.18 Describe how to carry out an acid-alkali titration, using burette, pipette and a suitable indicator, to prepare a pure, dry salt
• C5.4.7 describe and explain the procedure for a titration to give precise, accurate, valid and repeatable results
• 6 Titration of a strong acid and strong alkali to find the concentration of the acid using an appropriate pH indicator
• 7 Production of pure dry sample of an insoluble and soluble salt
• C5.1b describe the technique of titration
• PAG 6 Titration of a strong acid and strong alkali to find the concentration of the acid using an appropriate pH indicator
• C5.3.6 describe and explain the procedure for a titration to give precise, accurate, valid and repeatable results
• C4 Production of pure dry sample of an insoluble and soluble salt
• In an acid-base titration, the concentration of the acid or base is determined by accurately measuring the volumes used in the neutralisation reaction. An indicator can be added to show the end-point of the reaction
• Titration is used to determine, accurately, the volumes of solution required to reach the end-point of a chemical reaction.
• (j) titration as a method to prepare solutions of soluble salts and to determine relative and actual concentrations of solutions of acids/alkalis
• (f) acid-base titrations

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Acids and Bases

Learning objectives.

• Define equivalence point.
• Describe how to perform a titration experiment.
• Perform calculations to determine concentration of unknown acid or base.
• Describe titration curves of acid-base neutralization reactions.

Titration Experiment

Didn’t that used to be french fries.

A lot of research is going on these days involving the development of biodiesel fuels. Often this material can be made from used vegetable oils. The vegetable oil is treated with lye to create the biofuel. In the oils is a variable amount of acid that needs to be determined so the workers will know how much lye to add to make the final fuel. Before the lye is added, the native vegetable oil is titrated to find out how much free acid is present. Then the amount of lye added can be adjusted to take into account the amount needed to neutralize these free acids.

In the neutralization of hydrochloric acid by sodium hydroxide, the mole ratio of acid to base is 1:1.

One mole of HCl would be fully neutralized by one mole of NaOH. If instead the hydrochloric acid was reacted with barium hydroxide, the mole ratio would be 2:1.

Now two moles of HCl would be required to neutralize one mole of Ba(OH) 2 . The mole ratio insures that the number of moles of H + ions supplied by the acid is equal to the number of OH − ions supplied by the base. This must be the case for neutralization to occur. The equivalence point is the point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.

In the laboratory, it is useful to have an experiment where the unknown concentration of an acid or a base can be determined. This can be accomplished by performing a controlled neutralization reaction. A titration is an experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration. Many titrations are acid-base neutralization reactions, though other types of titrations can also be performed.

In order to perform an acid-base titration, the chemist must have a way to visually detect that the neutralization reaction has occurred. An indicator is a substance that has a distinctly different color when in an acidic or basic solution. A commonly used indicator for strong acid-strong base titrations is phenolphthalein. Solutions in which a few drops of phenolphthalein have been added turn from colorless to brilliant pink as the solution turns from acidic to basic. The steps in a titration reaction are outlined below.

• A measured volume of an acid of unknown concentration is added to an Erlenmeyer flask.
• Several drops of an indicator are added to the acid and mixed by swirling the flask.
• A buret is filled with the base solution of known molarity.
• The stopcock of the buret is opened and base is slowly added to the acid while the flask is constantly swirled to insure mixing. The stopcock is closed at the exact point at which the indicator just changes color.

Figure 1. Phenolphthalein in basic solution.

The standard solution is the solution in a titration whose concentration is known. In the titration described above the base solution is the standard solution. It is very important in a titration to add the solution from the buret slowly so that the point at which the indicator changes color can be found accurately.

The end point of a titration is the point at which the indicator changes color. When phenolphthalein is the indicator, the end point will be signified by a faint pink color.

Titration Calculations

How is soap made.

The manufacture of soap requires a number of chemistry techniques. One necessary piece of information is the saponification number. This is the amount of base needed to hydrolyze a certain amount of fat to produce the free fatty acids that are an essential part of the final product.

The fat is heated with a known amount of base (usually NaOH or KOH). After hydrolysis is complete, the left-over base is titrated to determine how much was needed to hydrolyze the fat sample.

At the equivalence point in a neutralization, the moles of acid are equal to the moles of base.

moles acid = moles base

Recall that the molarity ( M ) of a solution is defined as the moles of the solute divided by the liters of solution ( L ). So the moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.

We can then set the moles of acid equal to the moles of base.

Suppose that a titration is performed and 20.70 mL of 0.500 M NaOH is required to reach the end point when titrated against 15.00 mL of HCl of unknown concentration. The above equation can be used to solve for the molarity of the acid.

The higher molarity of the acid compared to the base in this case means that a smaller volume of the acid is required to reach the equivalence point.

The above equation works only for neutralizations in which there is a 1:1 ratio between the acid and the base. The sample problem below demonstrates the technique to solve a titration problem for a titration of sulfuric acid with sodium hydroxide.

Sample Problem: Titration

In a titration of sulfuric acid against sodium hydroxide, 32.20 mL of 0.250 M NaOH is required to neutralize 26.60 mL of H 2 SO 4 . Calculate the molarity of the sulfuric acid.

Step 1: List the known values and plan the problem.

• molarity NaOH = 0.250 M
• volume NaOH = 32.20 mL
• volume H 2 SO 4 = 26.60 mL
• molarity H 2 SO 4 = ?

First determine the moles of NaOH in the reaction. From the mole ratio, calculate the moles of H 2 SO 4 that reacted. Finally, divide the moles H 2 SO 4 by its volume to get the molarity.

Step 2: Solve.

Step 3: Think about your result.

The volume of H 2 SO 4 required is smaller than the volume of NaOH because of the two hydrogen ions contributed by each molecule.

Titration Curves

Where did graphs come from.

The  x–y  plot that we know of as a graph was the brainchild of the French mathematician-philosopher Rene Descartes (1596–1650). His studies in mathematics led him to develop what was known as “Cartesian geometry,” including the idea of our current graphs. The coordinates are often referred to as Cartesian coordinates.

As base is added to acid at the beginning of a titration, the pH rises very slowly. Nearer to the equivalence point, the pH begins to rapidly increase. If the titration is a strong acid with a strong base, the pH at the equivalence point is equal to 7. A bit past the equivalence point, the rate of change of the pH again slows down. A titration curve is a graphical representation of the pH of a solution during a titration. The Figure below shows two different examples of a strong acid-strong base titration curve. On the left is a titration in which the base is added to the acid and so the pH progresses from low to high. On the right is a titration in which the acid is added to the base. In this case, the pH starts out high and decreases during the titration. In both cases, the equivalence point is reached when the moles of acid and base are equal and the pH is 7. This also corresponds to the color change of the indicator.

Figure 2. A titration curve shows the pH changes that occur during the titration of an acid with a base. On the left, base is being added to acid. On the right, acid is being added to base. In both cases, the equivalence point is at pH 7.

Titration curves can also be generated in the case of a weak acid-strong base titration or a strong base-weak acid titration. The general shape of the titration curve is the same, but the pH at the equivalence point is different. In a weak acid-strong base titration, the pH is greater than 7 at the equivalence point. In a strong acid-weak base titration, the pH is less than 7 at the equivalence point.

Figure 3. Titration curve of weak acid and strong base.

• Definitions are given for equivalence point, titration and indicator.
• The process for carrying out a titration is described.
• The process of calculating concentration from titration data is described and illustrated.
• Acid-base titration curves are described.

Watch the video at the link below and answer the following questions:

• What is the indicator used?
• What color is it in acid solution?
• What is the glass tube called that contains the known concentration of sodium hydroxide?
• What other method can be used to determine the end-point of the titration?

Do the problems at the link below:

http://www.sophia.org/acidbase-titration-calculations-concept

Read the material at the link below and answer the following questions:

http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

• Why is the equivalence point less than pH 7 for the titration of ammonia with HCl?
• Why is it difficult to do a titration of a weak acid and a weak base?
• Why do we get two inflection points for the titration of ethanedioic acid?
• What is the standard solution?
• How do you know you have reached the end-point?
• What is the reaction that occurs during a titration?
• What assumption is made about the amounts of materials at the neutral point?
• What is different about the calculation using sulfuric acid?
• Why is the mole ratio important?
• What does a titration curve tell us?
• At what pH are the moles of acid and base equal?
• Is the equivalence point for a weak acid-strong base titration the same as for a strong-acid-strong base titration?
• end point: The point at which the indicator changes color.
• equivalence point: The point in a neutralization reaction where the number of moles of hydrogen ions is equal to the number of moles of hydroxide ions.
• indicator: A substance that has a distinctly different color when in an acidic or basic solution.
• standard solution: The solution in a titration whose concentration is known.
• titration: An experiment where a volume of a solution of known concentration is added to a volume of another solution in order to determine its concentration.
• titration curve: A graphical representation of the pH of a solution during a titration.
• User:GeorgHH/Wikimedia Commons. http://commons.wikimedia.org/wiki/File:Zapfs%C3%A4ule_044_3.jpg .
• Ben Mills (Wikimedia: Benjah-bmm27). http://commons.wikimedia.org/wiki/File:Phenolphthalein-at-pH-9.jpg .
• User:Phanton/Wikipedia. http://commons.wikimedia.org/wiki/File:Decorative_Soaps.jpg .
• William Holl the Younger (1807-1871) after Frans Hals. http://commons.wikimedia.org/wiki/File:William_Holl_the_Younger06.jpg .
• Laura Guerin. CK-12 Foundation.
• Chemistry Concepts Intermediate. Authored by : Calbreath, Baxter, et al.. Provided by : CK12.org. Located at : http://www.ck12.org/book/CK-12-Chemistry-Concepts-Intermediate/ . License : CC BY-NC: Attribution-NonCommercial

14.7 Acid-Base Titrations

Learning objectives.

By the end of this section, you will be able to:

• Interpret titration curves for strong and weak acid-base systems
• Compute sample pH at important stages of a titration
• Explain the function of acid-base indicators

As seen in the chapter on the stoichiometry of chemical reactions, titrations can be used to quantitatively analyze solutions for their acid or base concentrations. In this section, we will explore the underlying chemical equilibria that make acid-base titrimetry a useful analytical technique.

Titration Curves

A titration curve is a plot of some solution property versus the amount of added titrant. For acid-base titrations, solution pH is a useful property to monitor because it varies predictably with the solution composition and, therefore, may be used to monitor the titration’s progress and detect its end point. The following example exercise demonstrates the computation of pH for a titration solution after additions of several specified titrant volumes. The first example involves a strong acid titration that requires only stoichiometric calculations to derive the solution pH. The second example addresses a weak acid titration requiring equilibrium calculations.

Example 14.21

Calculating ph for titration solutions: strong acid/strong base.

(a) 0.00 mL

(b) 12.50 mL

(c) 25.00 mL

(d) 37.50 mL

(b) Titrant volume = 12.50 mL. Since the acid sample and the base titrant are both monoprotic and equally concentrated, this titrant addition involves less than a stoichiometric amount of base, and so it is completely consumed by reaction with the excess acid in the sample. The concentration of acid remaining is computed by subtracting the consumed amount from the intial amount and then dividing by the solution volume:

(c) Titrant volume = 25.00 mL. This titrant addition involves a stoichiometric amount of base (the equivalence point ), and so only products of the neutralization reaction are in solution (water and NaCl). Neither the cation nor the anion of this salt undergo acid-base ionization; the only process generating hydronium ions is the autoprotolysis of water. The solution is neutral, having a pH = 7.00.

(d) Titrant volume = 37.50 mL. This involves the addition of titrant in excess of the equivalence point. The solution pH is then calculated using the concentration of hydroxide ion:

pH = 14 − pOH = 14 + log([OH − ]) = 14 + log(0.0200) = 12.30

Check Your Learning

0.00: 1.000; 15.0: 1.5111; 25.0: 7; 40.0: 12.523

Example 14.22

Titration of a weak acid with a strong base.

Calculate the pH of the titration solution after the addition of the following volumes of NaOH titrant:

(b) 25.00 mL

(c) 12.50 mL

K a = [ H 3 O + ] [ CH 3 CO 2 − ] [ CH 3 CO 2 H ] ≈ [ H 3 O + ] 2 [ CH 3 CO 2 H ] 0 , K a = [ H 3 O + ] [ CH 3 CO 2 − ] [ CH 3 CO 2 H ] ≈ [ H 3 O + ] 2 [ CH 3 CO 2 H ] 0 , and [ H 3 O + ] = K a × [ CH 3 CO 2 H ] = 1.8 × 10 −5 × 0.100 = 1.3 × 10 −3 [ H 3 O + ] = K a × [ CH 3 CO 2 H ] = 1.8 × 10 −5 × 0.100 = 1.3 × 10 −3

(b) The acid and titrant are both monoprotic and the sample and titrant solutions are equally concentrated; thus, this volume of titrant represents the equivalence point. Unlike the strong-acid example above, however, the reaction mixture in this case contains a weak conjugate base (acetate ion). The solution pH is computed considering the base ionization of acetate, which is present at a concentration of

Base ionization of acetate is represented by the equation

Assuming x << 0.0500, the pH may be calculated via the usual ICE approach: K b = x 2 0.0500 M K b = x 2 0.0500 M

Note that the pH at the equivalence point of this titration is significantly greater than 7, as expected when titrating a weak acid with a strong base.

(c) Titrant volume = 12.50 mL. This volume represents one-half of the stoichiometric amount of titrant, and so one-half of the acetic acid has been neutralized to yield an equivalent amount of acetate ion. The concentrations of these conjugate acid-base partners, therefore, are equal. A convenient approach to computing the pH is use of the Henderson-Hasselbalch equation:

(pH = p K a at the half-equivalence point in a titration of a weak acid)

(d) Titrant volume = 37.50 mL. This volume represents a stoichiometric excess of titrant, and a reaction solution containing both the titration product, acetate ion, and the excess strong titrant. In such solutions, the solution pH is determined primarily by the amount of excess strong base:

0.00 mL: 2.37; 15.0 mL: 3.92; 25.00 mL: 8.29; 30.0 mL: 12.097

Performing additional calculations similar to those in the preceding example permits a more full assessment of titration curves. A summary of pH/volume data pairs for the strong and weak acid titrations is provided in Table 14.2 and plotted as titration curves in Figure 14.18 . A comparison of these two curves illustrates several important concepts that are best addressed by identifying the four stages of a titration:

initial state (added titrant volume = 0 mL): pH is determined by the acid being titrated; because the two acid samples are equally concentrated, the weak acid will exhibit a greater initial pH

pre-equivalence point (0 mL < V < 25 mL): solution pH increases gradually and the acid is consumed by reaction with added titrant; composition includes unreacted acid and the reaction product, its conjugate base

equivalence point ( V = 25 mL): a drastic rise in pH is observed as the solution composition transitions from acidic to either neutral (for the strong acid sample) or basic (for the weak acid sample), with pH determined by ionization of the conjugate base of the acid

postequivalence point ( V > 25 mL): pH is determined by the amount of excess strong base titrant added; since both samples are titrated with the same titrant, both titration curves appear similar at this stage.

Acid-Base Indicators

Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × × 10 −9 M (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × × 10 −9 M (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called acid-base indicators . Acid-base indicators are either weak organic acids or weak organic bases.

The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule:

The anion of methyl orange, In − , is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Châtelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers.

The perceived color of an indicator solution is determined by the ratio of the concentrations of the two species In − and HIn. If most of the indicator (typically about 60−90% or more) is present as In − , the perceived color of the solution is yellow. If most is present as HIn, then the solution color appears red. The Henderson-Hasselbalch equation is useful for understanding the relationship between the pH of an indicator solution and its composition (thus, perceived color):

In solutions where pH > p K a , the logarithmic term must be positive, indicating an excess of the conjugate base form of the indicator (yellow solution). When pH < p K a , the log term must be negative, indicating an excess of the conjugate acid (red solution). When the solution pH is close to the indicator pKa, appreciable amounts of both conjugate partners are present, and the solution color is that of an additive combination of each (yellow and red, yielding orange). The color change interval (or pH interval ) for an acid-base indicator is defined as the range of pH values over which a change in color is observed, and for most indicators this range is approximately p K a ± 1.

There are many different acid-base indicators that cover a wide range of pH values and can be used to determine the approximate pH of an unknown solution by a process of elimination. Universal indicators and pH paper contain a mixture of indicators and exhibit different colors at different pHs. Figure 14.19 presents several indicators, their colors, and their color-change intervals.

The titration curves shown in Figure 14.20 illustrate the choice of a suitable indicator for specific titrations. In the strong acid titration, use of any of the three indicators should yield reasonably sharp color changes and accurate end point determinations. For this titration, the solution pH reaches the lower limit of the methyl orange color change interval after addition of ~24 mL of titrant, at which point the initially red solution would begin to appear orange. When 25 mL of titrant has been added (the equivalence point), the pH is well above the upper limit and the solution will appear yellow. The titration's end point may then be estimated as the volume of titrant that yields a distinct orange-to-yellow color change. This color change would be challenging for most human eyes to precisely discern. More-accurate estimates of the titration end point are possible using either litmus or phenolphthalein, both of which exhibit color change intervals that are encompassed by the steep rise in pH that occurs around the 25.00 mL equivalence point.

The weak acid titration curve in Figure 14.20 shows that only one of the three indicators is suitable for end point detection. If methyl orange is used in this titration, the solution will undergo a gradual red-to-orange-to-yellow color change over a relatively large volume interval (0–6 mL), completing the color change well before the equivalence point (25 mL) has been reached. Use of litmus would show a color change that begins after adding 7–8 mL of titrant and ends just before the equivalence point. Phenolphthalein, on the other hand, exhibits a color change interval that nicely brackets the abrupt change in pH occurring at the titration's equivalence point. A sharp color change from colorless to pink will be observed within a very small volume interval around the equivalence point.

• 1 Titration of 25.00 mL of 0.100 M HCl (0.00250 mol of HCI) with 0.100 M NaOH.
• 2 Titration of 25.00 mL of 0.100 M CH 3 CO 2 H (0.00250 mol of CH 3 CO 2 H) with 0.100 M NaOH.

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13 Effect of Temperature and Solvent on Solubility

To evaluate the solubility of two solid solutes in two different solvents at different temperatures.

Expected Learning Outcomes

• Determine the solubility curve of substances.
• Explain the features of solubility curves based on intermolecular forces.

Textbook Reference

This experiment illustrates concepts from Tro, Chemistry: Structures and Properties , 2nd Ed., Ch. 13.2 and 13.4.

Introduction

The solubility  of a compound in a given solvent is the mass of solute that can be dissolved in a given amount of solvent. The solubility is typically expressed as

[latex]\textrm{solubility} = \frac{\textrm{g solute}}{100 \textrm{ g solvent}}[/latex]

When such a solution is formed, it is referred to as a saturated solution ; no more solute can be dissolved and additional solute will be suspended in the solution. [1]

The Solvation Process

The key to solubility is that, when a saturated solution is present, it is essentially the point where you are at an equilibrium between the suspension form and the solution form. This point is found when you start to see some specks of solid suspended in the liquid.

For sucrose in water, we have

\begin{equation} \textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11} (s) \rightleftharpoons \textrm{C}_{12}\textrm{H}_{22}\textrm{O}_{11} (aq) \end{equation}

For an ionic compound, as you may recall from CHEM-C 105 (Tro, Chemistry: Structures and Properties , 2nd Ed, Ch. 8), we need to account for the dissociation of the ionic compound.

Sodium chloride, NaCl, will dissociate when dissolved in water to form Na + and Cl – ions

\begin{equation} \textrm{NaCl}(s) \rightleftharpoons \textrm{Na}^+ (aq) + \textrm{Cl}^- (aq) \end{equation}

The solvation process can be considered as a three-step process:

• Breaking the solute-solute interactions
• Breaking the solvent-solvent interactions.
• Forming solvent-solute interactions – attraction between the solute and solvent particles.

The first two processes above are endothermic while the last process is exothermic.   The overall thermodynamics is summarized in the following enthalpy level diagram; note that the overall enthalpy of solvation

In this experiment, you will study the solubility of potassium dichromate ([latex]\textrm{K}_2\textrm{Cr}_2\textrm{O}_7[/latex]) or oxalic acid ([latex]\textrm{H}_2\textrm{C}_2\textrm{O}_4[/latex]) using two different solvents:

• 70:30 water:1,4-dioxane (by volume) mixture

1,4-dioxane is (largely) nonpolar due to its symmetrical shape. The structures of both oxalic acid and 1,4-dioxane are illustrated below:

Effect of Temperature on Solubility

The temperature dependence of the solubility of substances are depicted graphically on a solubility curve.

For many solid solutes in liquid solvents (as we see from everyday life) the solubility of the solute increases with temperature.  However, this is not a hard and fast rule.  For example, if you look at the solubility curve of cadmium selenate above, the solubility decreases as a function of temperature.  As you could also see above, how much the solubility changes as a function of temperature varies significantly for different substances.

In this experiment, you will explore this for both of the solids that are studied in this experiment in both water and the water:1,4-dioxane mixture. Wwe will explore this in a more quantitative manner in the experiment  Thermodynamics of the Solvation of Calcium Hydroxide .

Solubility Differences of a Solute in Different Solvents

As discussed above, the energetics of the solvation process involves a consideration of the solute-solute , solvent-solvent , and solute-solvent  interactions. While entropic considerations mean that exothermic interactions overall are not necessary for solvation, it does mean that solvation is unlikely unless the solute-solvent interactions are comparable (or larger) than the solute-solute and solvent-solvent interactions.

Sodium chloride forms ion-dipole forces with water, which (mostly) counterbalances the ionic bonds (within sodium chloride) and hydrogen bonds (within water) that are broken. As a result, sodium chloride is soluble in water.

On the other hand, sodium chloride is insoluble in cyclohexane (C 6 H 12 ), a non-polar solvent.  This is because the energy required to break the ionic bonds in sodium chloride is much greater than the gain in van der Waals energy when sodium and chloride ions interact with cyclohexane molecules.

A common way of thinking about this is to use the like dissolves like approach (which largely holds though there are subtle nuances to be considered).

• Ionic/polar solutes are more likely to be soluble in polar solvents
• Non-polar solutes are more likely to be soluble in non-polar solvents.

In this experiment, you will compare the solubility of each of the two solutes in the two solvent systems studied, and evaluate the difference in the context of this discussion.

In this experiment, you will be assigned to measure the solubility curves for either potassium dichromate  or oxalic acid.  You will then (before leaving the lab) share data with another group of students who did the measurements for the other solute.

• Volumes of solvent can be measured using a 5 mL graduated pipet.
• Part C must be completed in a fume hood.
• Students may be asked to complete the experiment in a different order from that listed here to help traffic control. This will not affect the results of the experiment.
• Throughout this experiment, it is important to stir the test tube continuously in a gentle manner such that the temperature throughout the test tube. On the other hand, you must do it in a manner such that you don’t break the test tube.

Part A: Solubility of a Solid Solute in Water

• On a piece of weighing paper, weigh out (as assigned) either 3.1-3.3 g oxalic acid or 2.8-2.9 g potassium dichromate.  Be sure to record the exact mass of your solute.
• Add the solid into a medium sized test tube.
• Using a graduated pipet, add 5.0 mL deionized water into the test tube. Use a test tube holder to clamp the test tube.
• Place the test tube into a 400 mL beaker of warm tap water. Begin heating the beaker on a hot plate, using the beaker as a water bath.
• Stir the mixture in the test tube regularly using the thermometer, keeping the test tube in the water bath until all of the solid has dissolved. You may also wish to stir (using a glass rod) the beaker of hot water from time to time.
• When all of the solid has dissolved, take the test tube out of the hot water beaker. Continue stirring the test tube gently while the test tube cools.
• Record the temperature when you first see for certain crystals of solute come out of solution.  Be careful not to confuse contaminants (e.g. specks of dust) with the solute.
• Add 3.0 mL deionized water into the test tube. Place the test tube back into the hot water beaker and repeat steps 5-7.
• Add 2.0 mL deionized water into the test tube. Place the test tube back into the hot water beaker and repeat steps 5-7.

Part B: Solubility of Your Solute in a Dilute Solution

• Prepare a 400 mL beaker containing ice. You may wish to add some salt to the ice as well.
• Weigh out and place into a clean, dry test tube approximately 0.9 g of your assigned solid.
• Add 5.0 mL of deionized water into the test tube.
• Repeat steps 5-7 from Part A.  If the solute is still completely dissolved at room temperature, place the test tube into the ice/salt bath and allow the mixture to cool until crystals of your solid are observed.
• Add 2.0 mL deionized water into the test tube and repeat step 14 twice.

Part C: Solubility of Your Solute in a Water:1,4-Dioxane Mixture

• 1.4-1.6 g potassium dichromate
• 3.2-3.5 g oxalic acid
• Move your test tube, graduated pipet, two 400 mL beakers (one containing your ice bath), thermometer, and test tube holder to a space in the fume hood.
• Measure out 5.0 mL of the 70:30 (v/v) water:dioxane mixture into the test tube.
• Repeat step 14 from Part B above.
• Add 3 mL of the 70:30 (v/v) water:dioxane mixture to the test tube and repeal step 14 from Part B again.
• Repeat step 20 two more times.
• Be sure to obtain the experimental data on the solid you did not study from another pair of students before you leave the laboratory.

Waste Management

All waste must be collected and discarded into appropriate beakers placed in the fume-hood.  There will be two separate waste beakers: one for waste containing oxalic acid and one for waste cotnaining potassium dichromate.

Data Analysis

For this experiment, you will need to first determine the solubility of the solid for each trial.  Since the volume of the solvent was measured at room temperature, the density value used should be that at room temperature (25°C):

Using these density values, calculate the mass of solvent used for each data point in this experiment.  Note that the volume of solvent used should be the total volume of solvent added, not the amount added at that point.

If you have first added 10.0 mL water, then added another 5.0 mL water, the total volume of solvent added is [latex]10.0 \textrm{ mL} + 5.0 \textrm{ mL}= 15.0\textrm{ mL}[/latex].  You will then calculate the mass of the solvent as:

[latex]?\textrm{ g} = 15.0\textrm{ mL} \times \frac{0.9975\textrm{ g}}{\textrm{mL}} = 14.96 \textrm{ g}[/latex]

From this, calculate the concentration of the solute (in g solute/(100 g solvent)) for each data point.

If the mass of solute is 5.912 g (should be the same for the entire part) and for that given trial you had 15.0 mL water (and hence the mass of solvent is 14.96 g from above), the concentration of this solution is

\begin{eqnarray} ? \frac{\textrm{g solute}}{100\textrm{ g solvent}} &=& 100\mbox{ g solvent} \times \frac{5.912\textrm{ g K}_2\textrm{Cr}_2\textrm{O}_7}{9.98\textrm{ g H}_2\textrm{O}} \\ &=& 59.3\textrm{ g solute/100 g solvent} \end{eqnarray}

Make two plots (one for each solute) where you plot the solubility in each solvent (along the  y -axis) as a function of temperature (along the x -axis).

On each graph, there should be two data sets along the same axes.

• The solute in water (Parts A and B). The data for both those parts (for a given solute) should fall along a single, smooth curve.
• The solute in a 70:30 water:1,4-dioxane mixture (Part C). The data for this part (for a given solute) should fall on a separate, smooth curve.

Each set should be plotted with different symbols, and a smooth curve to illustrate the trend in the data (as best as possible) should be included to guide the eye for each set as shown in the illustration below.  It is common for there to be anomalous data points in this data, so you should not expect the curve to go through every data point or for every “jump” to follow the overall trend.

• Well, there are supersaturated solutions where there is a greater amount of solute dissolved than what is found in the solubility. However, such solutions are not thermodynamically stable and will not be considered in this experiment. ↵

IU East Experimental Chemistry Laboratory Manual Copyright © 2022 by Yu Kay Law is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License , except where otherwise noted.

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AP®︎/College Chemistry

Course: ap®︎/college chemistry   >   unit 7, introduction to solubility equilibria.

• Worked example: Calculating solubility from Kₛₚ
• Worked example: Predicting whether a precipitate forms by comparing Q and Kₛₚ
• The common-ion effect
• pH and solubility
• Solubility equilibria

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Errors in Titration Experiments

Reasons for Error in a Chemistry Experiment

Titration is a sensitive analytical method that lets you determine an unknown concentration of a chemical in solution by introducing a known concentration of another chemical. Several factors can cause errors in titration findings, including misreading volumes, mistaken concentration values or faulty technique. Care must be taken as the solution of the known concentration is introduced into a specific volume of the unknown through laboratory glassware such as a burette or pipette. Indicators are used to determine when a reaction has come to an end.

End Point Error

The end point of a titration is when the reaction between the two solutions has stopped. Indicators, which change color to indicate when the reaction has stopped, do not change instantly. In the case of acid-base titration, the indicator may first lighten in color before changing completely. Also, each individual perceives color slightly differently, which affects the outcome of the experiment. If the color has changed slightly, too much of the titrant, which comes from the burette, can be introduced into the solution, overshooting results.

Misreading the Volume

The accuracy of titration requires precise measurement of the volume of materials in use. But markings on a burette can be easily misread. One way to misread the volume is by looking at the measurement on an angle. From above, it can seem like the volume is lower, while from below, the apparent volume looks higher. Another source of measurement error is looking at the wrong spot. A solution forms a concave curve and the bottom of the curve is used to measure the volume. If the reading is taken from the higher sections of the curve, the volume measurement will be in error.

Concentrations

Errors in concentrations directly affect the measurement accuracy. Errors include using the wrong concentration to begin with, which can occur from chemical decomposition or evaporation of fluids. The solution may have been prepared incorrectly or contaminatns could have been introduced into the solution, such as using dirty equipment. Even the process of cleaning your equipment, if carried out with the wrong solution, can affect the concentrations of the solutions to be experimented on.

Using the Equipment Incorrectly

You must follow strict guidelines in handling and using all equipment during the experiment as the slightest mistake can create errors in the findings. For example, swirling the solution can result in loss of solution that will affect results. Errors in filling the burette can cause air bubbles that affect the flow of the liquid in the burette.

Other Errors

Other human or equipment errors can also creep in. Human error includes using selecting the wrong reagents or using the wrong amount of indicator. Equipment error typically is in the burette, which can develop leaks over time. Even a small loss of fluid will affect the results of the titration.

Related Articles

Purpose of titration, laboratory glassware and functions, how to make dilutions, how to know when a titration is complete, use of titration, how to determine the concentration of a titration, what is "direct titration", how to write a lab report about titration, titration of sodium carbonate with hydrochloric acid, what does titration mean, precipitation titration techniques, titration explained, how to calculate the pka in titration, how to standardize a ph meter, acid base titration theory, acid base titration sources of error improvements, how to calculate the k value on a titration graph.

• Titrations.info: Titration Errors
• Purdue: What Is a Titration?

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