Question

$MY$ and $NY_3$, two nearly insoluble salts, have the same $K_{SP}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $MY$ and $NY_3$ ?

• A. The molar solubility of MY in water is less than that of $NY_3$
• B. The salts $MY$ and $NY_3$ are more soluble in 0.5 M KY than in pure water
• C. The addition of the salt of KY to solution of $MY$ and $NY_3$ will have no effect on their solubilities
• D. The molar solubilities of $MY$ and $NY_3$ in water are identical

Answer 1

The Correct Option is A

 To determine which statement is true regarding the molar solubility of \(MY\) and \(NY_3\), given that both have the same solubility product constant \(K_{SP} = 6.2 \times 10^{-13}\), we need to understand the relation between solubility product and molar solubility of salts.

Dissolution Reactions:

• For \(MY\): \(MY \rightleftharpoons M^+ + Y^−\)
• For \(NY_3\): \(NY_3 \rightleftharpoons N^{3+} + 3Y^−\)

Expression for \(K_{SP}\):

• For \(MY\): \(K_{SP} = [M^+][Y^-] = s^2\), where \(s\) is the molar solubility of \(MY\).
• For \(NY_3\): \(K_{SP} = [N^{3+}][Y^-]^3 = s(3s)^3 = 27s^4\), where \(s\) is the molar solubility of \(NY_3\).

Calculating Molar Solubility:

• For \(MY\), let the molar solubility be \(s_1\): \(s_1^2 = 6.2 \times 10^{-13}\)
• Solving for \(s_1\): \(s_1 = \sqrt{6.2 \times 10^{-13}} = \approx 7.87 \times 10^{-7}\)
• For \(NY_3\), let the molar solubility be \(s_2\): \(27s_2^4 = 6.2 \times 10^{-13}\)
• Solving for \(s_2\): \(s_2^4 = \frac{6.2 \times 10^{-13}}{27}\)
• \(s_2 = \sqrt[4]{2.3 \times 10^{-14}} \approx 1.82 \times 10^{-4}\)

Therefore, the molar solubility of \(MY\) in water is less than that of \(NY_3\) as \(7.87 \times 10^{-7} < 1.82 \times 10^{-4}\).

The correct statement is: "The molar solubility of \(MY\) in water is less than that of \(NY_3\)".