The Common Ion Effect and Solubility (2024)

Introduction:

Potassium hydrogen tartrate (cream of tartar), KHC₄H₄O₆, is a weak acid, that is not very soluble in water.Its solubilityequilibrium in water is:

KHC₄H₄O₆ (s) K⁺(aq) + HC₄H₄O₆^-(aq)

The HC₄H₄O₆^-(aq) ion contains one acidichydrogen, so that the quantity of potassium hydrogen tartrate in solution can bedetermined by titrationwith a base. Because of the complexity of the formula, we will refer to the KHC₄H₄O₆as KHT in order to simplify the equation. Thus, we can write the solubilityreaction as:

KHT (s) K⁺(aq) + HT^-(aq)

The HT^-ions react as a monoprotic acid when titrated with NaOH.

HT^- (aq) + OH^- (aq) H₂O + T^2-(aq)

From these equations we can see that there will be a one to one relationship betweenthe moles of OH^- that are used in the titration, and the amount of KHT thatdissolves.

Now, what will happen if we add some KCl to the saturated solution? From leChâtelier's principle we can predict that the reaction will respond by trying to removesome of the added K⁺ ions.

Applied Stress	Le Châtelier's Principle Prediction of Response to Stress
	KHT (s)	K⁺ (aq)	+ HT^- (aq)
Increase [K⁺ (aq)] by adding KCl (s)		Decrease

The net result will be that there is a shift to the left, so the solubility decreases.

Applied Stress	Le Châtelier's Principle Prediction of Response to Stress
	KHT (s)	K⁺ (aq)	+ HT^- (aq)
Increase [K⁺ (aq)] by adding KCl (s)	Increase	Decrease	Decrease

The amount of additional KHT is too small to see; however, by measuring the actualamount of HT^- by titration, we can get the quantitative change in solubilitydue to the added common ion, K⁺.

Materials:

buret
50 mL graduated cylinder
about 150 mL of KCl solution (0.05, 0.10, 0.20 or 0.40 M, as assigned), or de-ionized water
standardized solution of NaOH (approximately 0.0500 M)
filter paper
funnel
Erlenmeyer flask
2 small beakers
phenolphthalein indicator
2 500 ml PET soda bottles

Procedures:

Pure potassium hydrogen tartrate (KHT) is not dangerous; in fact, it isone of the ingredients in baking powder. However, in the lab it may be contaminatedwith other substances, so treat it as if it were dangerous. While the concentrationof NaOH used in this experiment is not high, sodiumhydroxide is extremely corrosive. Make sure to wear your safety goggles at alltimes. A lab apron is also a good idea to avoid staining your clothes.

1. Each person in each lab group will prepare a saturated solution of KHT in pure wateror an assigned concentration of KCl. Prepare the saturatedsolutions of KHT as follows (this will give you enough for two titrations if you make a mistake you will have to prepare more solution):

add about 1 g of KHT to the PET bottle
put about 150 mL of water (or the assigned KCl solution) into the bottle
cap, shake intensely and vigorously for 5 minutes

2. Fold a filter paper into halves, then into quarters. Place this filter paperinto a dry funnel. Pour in some of your saturated solution of KHT, and filter about10 mL into a clean graduated cylinder. The first few mL should be used to wash outthe cylinder and thrown away. Then filter exactly 50 mL of the solution into thegraduated cylinder, and pour it into the Erlenmeyer flask. Add one or two drops ofphenolphthalein indicator.

3. Rinse the buret with a few mL of the standardized NaOH solution. Then fill theburet and measure the initial volume. Titrate the KHT solution to a permanent pinkendpoint that lasts for at least 10 seconds.

4. Repeat steps 2 and 3 with another 50 mL of KHT solution. You will now have twosets of data from which you can calculate the molar concentration of KHT.

Conclusion:

1. Calculate the [HT^-] from your titration data([HT^-] is the unknown concentration of the acid) for each trial. Calculate the average for the two trials.

2. Using the balanced equation, indicate why the [HT^-] is the same as thesolubility of KHT (s).

3. Tabulate the data for the entire class. Compare the solubility of KHT (s) asthe concentration of K⁺ (aq) ions increases from the dissolved KCl. Explain your results.

Extension:

Using a spreadsheet if possible, calculate the total concentration of K⁺(aq) ions in each solution. To do this remember that there are two sources of K⁺(aq):

from the dissolved KHT (so this [K⁺ (aq)] = [HT^-] )
from the dissolved KCl (so this [K⁺ (aq)] = [KCl] )

so that the [K⁺ (aq)]_total = [HT^-] + [KCl]

Substitute this value of [K⁺(aq)]_total into the K_spexpression and calculate K_sp for each solution used (from pure water to 0.40 MKCl).

Plot a graph of K_sp as a function of [K⁺(aq)]_total. Fit it to a linear graph, and extrapolate back to [K⁺(aq)]_total= 0. What is the value of K_sp at this point?

You will notice when you do this that Ksp is not constant, but in fact changes with [K⁺(aq)]_total. This is because we have used concentrations tocalculate the Ksp, and we should really use activities, which is a measure of how reactivea substance is. Only in very dilute solutions as we extrapolate back to 0 concentration does theactivity equal the concentration.

In order for an ion to dissolve in water it must cause some ordering or structure inthe water molecules. This is because the positive ends of water molecules (the Hatoms) are going to be attracted towards any negative ions, and the negative ends (the Oatoms) to any positive ions. The greater the amount of ions present in the water,the more structured the water will be. This will make it easier for other ions todissolve. This polarity effect enhances solubility, so in general, non-reacting ionscause an increase in activity (and thus in solubility) of ionic solids. The greaterthe concentration of ions, the more this effect takes place, and so the greater theactivity of the ions becomes.

So, for potassium hydrogen tartrate KHT (s) dissolving in water:

KHT (s) K⁺(aq) + HT^-(aq)

the equilibrium constant where is the activity of the K⁺ions and is the activity of the HT^-ions. Since activity increases with increasing concentration, it is obvious that sodoes the K_sp

Common ions of course are different, since they will take part in the reaction. Acommon ion will change the equilibrium as predicted by le Châtelier's principle. However, the common ion will always have a non-reacting oppositely charged ion with it, soit too will have an impact on the activity.

You can demonstrate this effect in the experiment to measure the solubility ofpotassium hydrogen tartrate (KHT). Simply add some NaCl (which has no ions in commonwith the KHT), and the solubility of the KHT will increase.