The solubility of a slightly soluble salt can be greatly affected by the additio
ID: 524213 • Letter: T
Question
The solubility of a slightly soluble salt can be greatly affected by the addition of a soluble salt with a common ion, that is, with one of the ions in the added soluble salt being identical to one of the ions of the slightly soluble salt. The general result of the addition of the common ion is to greatly reduce the solubility of the slightly soluble salt. In other words, the addition of the common ion results in a shift in the equilibrium of the slightly soluble salt.
Part A
Mg(OH)2 is a sparingly soluble salt with a solubility product, Ksp, of 5.61×1011. It is used to control the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams. What is the ratio of solubility of Mg(OH)2dissolved in pure H2O to Mg(OH)2dissolved in a 0.140 M NaOH solution?
Express your answer numerically as the ratio of molar solubility in H2O to the molar solubility in NaOH.
A buffer is a mixed solution of a weak acid or base, combined with its conjugate. Note that this can be understood essentially as a common-ion problem: The conjugate is a common ion added to an equilibrium system of a weak acid or base. The addition of the conjugate shifts the equilibrium of the system to relieve the stress of the added concentration of the common ion. In a solution consisting of a weak acid or base, the equilibrium shift also results in a pH shift of the system.
It is the presence of the common ion in the system that results in buffering behavior, because both added H+ or OH ions can be neutralized.
Part B
What is the pH change of a 0.300 Msolution of citric acid (pKa=4.77) if citrate is added to a concentration of 0.130 M with no change in volume?
Express the difference in pH numerically to two decimal places.
Explanation / Answer
Ksp = 5.61*10^-11
if [NAOh] = 0.14 M then, find Solubility ratio
S water / S solution
Swater -->
Ksp = [Mg+2][OH-]^2
5.61*10^-11 = (S)(2S)^2
4*S^3 = 5.61*10^-11
S = ((5.61*10^-11)/4)^(1/3) = 0.00024115 M
in [NaOH] = 0..14 M
[OH-] = 0.14 M
Ksp = [Mg+2][OH-]^2
5.61*10^-11 = (S)(0.14^2)
S = (5.61*10^-11)/((0.14^2)) = 2.8622*10^-9 M
Ratio = 0.00024115 /(2.8622*10^-9)
Ratio = 84253.371
B)
pH change if we add
pH = pKa + log(citrate/citric acid)
pH = 4.77 + log(0.130 / (0.3 -0.130 ))
pH = 4.65