What is the solubility (in grams/L) of lead (II) chromate (PbCrO4, molar mass = 323 g/mol, Ksp = 2*10^-14) in 0.45 M K2CrO4
A 0.350 M solution of the weak acid Ha has pH 4.35 at 25 degrees C. What is the △G for the dissociation of the weak acid?
The Correct Answer and Explanation is :
Problem 1: Solubility of PbCrO₄ in 0.45 M K₂CrO₄
The solubility equilibrium of lead(II) chromate is:
[
PbCrO_4 (s) \rightleftharpoons Pb^{2+} (aq) + CrO_4^{2-} (aq)
]
The solubility product constant is:
[
K_{sp} = [Pb^{2+}][CrO_4^{2-}]
]
Let ( s ) be the solubility of PbCrO₄. Since the solution contains 0.45 M ( K_2CrO_4 ), the chromate ion concentration is initially 0.45 M before any dissolution.
Since dissolution adds ( s ) amount of ( Pb^{2+} ) and ( s ) amount of ( CrO_4^{2-} ), the equilibrium expression becomes:
[
K_{sp} = (s)(0.45 + s)
]
Given ( K_{sp} = 2.0 \times 10^{-14} ), and since ( s ) is very small compared to 0.45, we approximate:
[
K_{sp} \approx (s)(0.45)
]
Solving for ( s ):
[
s = \frac{2.0 \times 10^{-14}}{0.45} = 4.44 \times 10^{-14} \text{ M}
]
To convert to g/L:
[
\text{Solubility} = (4.44 \times 10^{-14} \text{ mol/L}) \times (323 \text{ g/mol}) = 1.43 \times 10^{-11} \text{ g/L}
]
Problem 2: Gibbs Free Energy for Weak Acid Dissociation
The dissociation of the weak acid ( HA ) follows:
[
HA \rightleftharpoons H^+ + A^-
]
Given:
- ( [HA] = 0.350 ) M
- ( \text{pH} = 4.35 ), so ( [H^+] = 10^{-4.35} = 4.47 \times 10^{-5} ) M
Using the dissociation constant formula:
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Since equal amounts of ( H^+ ) and ( A^- ) are formed,
[
K_a = \frac{(4.47 \times 10^{-5})^2}{0.350} = 5.71 \times 10^{-9}
]
To find Gibbs free energy:
[
\Delta G = -RT \ln K_a
]
Using ( R = 8.314 ) J/(mol·K) and ( T = 298 ) K:
[
\Delta G = – (8.314)(298) \ln (5.71 \times 10^{-9})
]
[
= – (8.314)(298) (-18.38)
]
[
= 45.6 \text{ kJ/mol}
]
Thus, ( \Delta G \approx 45.6 ) kJ/mol.
Explanation:
Gibbs free energy change ((\Delta G)) indicates whether a reaction is spontaneous. A positive ( \Delta G ) (as we found here) means the dissociation of this weak acid is non-spontaneous under standard conditions. The equilibrium constant (( K_a )) is quite small, which means the acid does not ionize significantly in solution. This aligns with our large positive ( \Delta G ), indicating that the reverse reaction (association of ( H^+ ) and ( A^- )) is favored over dissociation.