AP CHEMISTRY LAB
DETERMINATION OF THE SOLUBILITY-PRODUCT CONSTANT
FOR CREAM OF TARTAR
THEORY:
From our study of the solubility rules, you will remember that silver will form a precipitate
when it is mixed with the chromate ion. Specifically, a molecular equation for a reaction between
silver nitrate and potassium chromate is: 2AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + 2KNO3(aq).
Since Ag2CrO4(s) is the precipitate, or slightly soluble salt, in the reaction, we have an equilibrium
system set up with this salt. From our knowledge of equilibrium constants, we should be able to set
=
up the equation and equilibrium constant expression for this salt: Ag 2 CrO4( s ) ↔ 2Ag (+aq ) + CrO 4(
aq ) ,
then the Ksp = [Ag + ]2 [CrO=4( aq ) ] . RT has had his students determine that value for several years
with varying degrees of success. Today, we are going to look at another compound and it solubility,
cream of tartar or potassium hydrogen tartrate. Both of those names are rather long, so the name will
be shortened to KHT.
KHT is an interesting and commercially relevant compound. “The preparation of a good
wine often involves a long period of storage at 25 to 28o F. The cooling procedure decreases the
solubility of KHT in the wine and the resulting crystals of KHT are filtered out. This helps to
prevent formation of haziness in the final wine produce that can sometimes result from
precipitation of the KHT. The decrease in KHT concentration also often helps to adjust the
acidity of the wine to a proper level. The dependence of KHT concentration on temperature
could be determined. However, that is a time consuming process, so we will forgo that process.
That data could be used to calculate the solubility product of KHT.” We will also see the effect
of a common ion on the KHT concentration.
There are three factors that need to be considered here. First, since the cloudiness of the
solution will be determined throughout most of the lab by observing it with our eyes, the
procedure has inherent errors in it. We should, however, be able to get a Ksp value that is
accurate to the first order of magnitude. Secondly, we will be adding a common ion, K+, to the
saturated solution at room temperature to see the result of the solubility of the salt. (We will not
preform the first 2 considerations because of time constrants!) Finally, to be more precise for
our determination of the Ksp value, we will titrate the solution. This method will work since
KHT is an acid salt. Two equations are relevant at this time. First the dissociation reaction for
−
KHT: KHC4 H 4 O6( s ) → K (+aq ) + HC4 H 4O6(
aq ) . The second equation is:
HC4 H 4 O−6( aq ) + NaOH →[NaC4 H 4O6( aq ) ]− + H 2O(l ) .
The general idea a sample of KHT will be prepared and a measured volume of that
saturated solution will be titrated with a standard NaOH solution. From that data the
[HC4 H 4O6(− aq ) ] can be determined. By stoichiometry, we will be able to determine the K (+aq ) and
then the Ksp for the slat. It’s time to get started!
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PURPOSE:
To become familiar with equilibrium involving potassium hydrogen tartrate.
To practice titration techniques.
Observe the result of LeCharelier’s Principle for a common ion. (Will be demonstrated in
class)
Reinforce techniques learned to prepare standard solutions.
PROCEDURE:
1) RT has prepared your KHT solution for the titration part of the lab. Pipette two 20.00 ml samples
of the KHT solution into two clean 250 ml Erlenmeyer flasks.
2) Titrate your samples using 2 drops of phenolphthalein as your indicator. Make sure that you
record the concentration of the NaOH solution that you used as the titrant. (Remember, earlier in
the year, we standardized a NaOH solution. Hopefully, you have that concentration in you
notebook. It should be close to 0.10 M.)
CALCULATIONS:
Before you do the lab:
1) Write the dissociation reaction for KHT.
2) Write the symbolic expression for the Ksp for KHT.
3) Write the molecular equation for the reaction between KHT and NaOH.
4) Write the net ionic equation for the reaction between KHT and NaOH.
5) Look up and record the Ksp value for KHT in a CRC or Lang’s Hand book of Chemistry. You
may have to work with the solubility of the salt to calculate the Ksp value. This value may be
found on the web also.
After the lab:
6) Show sample calculations for:
A) Moles of KHT.
B) Concentration of one of your samples of KHT.
C) How you calculate Ksp for one of your sample. Make sure you put in the
symbolic expression before you do the calculation.
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D) The [HC4 H 4O6(
aq ) ] you found through titration and the resulting Ksp.
7) Collect the titration values for the Ksp of KHT form your classmates and find the class average.
Using your average determine the + value for your value.
8) If you were able to find the accepted value for the Ksp of KHT, find your % error for your titration
value.
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OTHER QUESTIONS & CALCULATIONS:
9) Calculate:
A) The moles of Ag+ in 5.0 ml of 0.0040 M AgNO3.
B) The number of moles of CrO4= in 5.0 ml of 0.0024 M K2CrO4.
C) Q if the solutions in 10 A & B are mixed.
D) Determine if a precipitate will form in the solution you prepared in 10 C.
(Ksp value for Ag2CrO4 is 1.1X10-12)
11) Given the Ksp for BaCrO4 = 1.09X10-5, will BaCrO4 precipitate when 10.0 ml of
1X10-4 M Ba(NO3)2 is mixed with 10 ml of 1X10-4 M K2Cr04?
12) The Ksp value for Ag2CrO4 is 1.1X10-12 and the Ksp for BaCrO4 is 1.09X10 -5. If a 0.025 M
solution of K2CrO4 was slowly added to a solution that contained 5.0X10-4 M of Ba2+ and
Ag+, which chromate salt would precipitate first? Justify your answer with calculations.
A reference: http://faculty.kutztown.edu/vitz/limsport/LabManual/KSPWeb/KSP.htm
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