Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 4.36. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.
To determine the molar solubility (S) of Zn(CN)2, we need to consider the effect of pH on the solubility of the compound. Here's a step-by-step approach to solving the problem:
Step 1: Write the balanced equation for the dissociation of Zn(CN)2 in water.
Zn(CN)2 ⇌ Zn2+ + 2CN-
Step 2: Write the expression for the solubility product (Ksp) of Zn(CN)2.
Ksp = [Zn2+][CN-]²
Step 3: Determine the concentrations of Zn2+ and CN- in terms of the molar solubility (S) of Zn(CN)2.
[Zn2+] = S
[CN-] = 2S (since the stoichiometry is 1:2)
Step 4: Substitute the concentrations into the expression for Ksp.
Ksp = S * (2S)² = 4S³
Step 5: Determine the value of S by solving for S in terms of Ksp.
S = ∛(Ksp/4)
Step 6: Calculate the value of Ksp.
Ksp = 3.0 × 10⁻¹⁶
Step 7: Substitute the value of Ksp into the equation for S.
S = ∛(3.0 × 10⁻¹⁶/4)
Step 8: Evaluate the expression to find the molar solubility of Zn(CN)2.
S ≈ 3.0 × 10⁻⁶
Therefore, the molar solubility of Zn(CN)2 in a solution with a pH = 4.36 is approximately 3.0 × 10⁻⁶ mol/L.
To determine the molar solubility (S) of Zn(CN)2 in a solution with a pH of 4.36, we need to consider the dissociation of HCN, as it can react with Zn(CN)2 and affect its solubility.
First, let's write the balanced equation for the dissociation of HCN in water:
HCN + H2O ⇌ H3O+ + CN-
From the given information, we know the equilibrium constant (Ka) for this reaction, which is 6.2 × 10^-10. Ka is the ratio of the products (H3O+ and CN-) to the reactant (HCN), so we can set up an expression for it:
Ka = [H3O+][CN-] / [HCN]
Since the pH is given as 4.36, we can use the relationship: pH = -log[H3O+]. Taking the negative logarithm of both sides and rearranging, we find:
[H3O+] = 10^(-pH)
Substituting this into the equation for Ka, we have:
Ka = [CN-] × 10^(-pH) / [HCN]
Since Zn(CN)2 will dissociate into Zn2+ and 2 CN- ions, we can write the solubility product expression (Ksp) for Zn(CN)2 as:
Ksp = [Zn2+][CN-]^2
We are given that Ksp (Zn(CN)2) = 3.0 × 10^-16.
To determine the molar solubility of Zn(CN)2 (S), we need to find the concentration of CN- ions. Since CN- is common to both the Ka and Ksp expressions, we can set them equal to each other:
[Zn2+][CN-]^2 = [CN-] × 10^(-pH) / [HCN]
Dividing both sides by [CN-], we get:
[Zn2+][CN-] = 10^(-pH) / [HCN]
Since we know the concentration of HCN, we can substitute the given values to calculate the molar solubility:
S = [CN-] = (10^(-pH) / [HCN]) / [Zn2+]
Substituting the given values of pH and the equilibrium constant Ka, as well as any known concentration of HCN, will give you the molar solubility (S) of Zn(CN)2 in the solution.