Determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 4.91. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.
3.35E-4
To determine the molar solubility of Zn(CN)2, we need to consider the pH of the solution and the solubility product constant (Ksp) of Zn(CN)2.
First, let's write the balanced chemical equation for the dissociation of Zn(CN)2 in water:
Zn(CN)2 ⇌ Zn2+ + 2CN-
Since Zn(CN)2 dissociates into Zn2+ and CN- ions, we need to determine the concentration of these ions in the solution.
Let's assume that the molar solubility of Zn(CN)2 is represented by 'S'. Therefore, the concentration of Zn2+ ions will be 'S' M and the concentration of CN- ions will be '2S' M.
Given that the solubility product constant (Ksp) of Zn(CN)2 is 3.0 × 10-16, we can write the following equation:
Ksp = [Zn2+][CN-]^2
Substituting the concentrations:
3.0 × 10-16 = (S)(2S)^2
Simplifying the equation:
3.0 × 10-16 = 4S^3
Dividing both sides by 4:
7.5 × 10-17 = S^3
Taking the cube root of both sides:
S ≈ 2.21 × 10-6 M
Therefore, the molar solubility (S) of Zn(CN)2 in a solution with a pH of 4.91 is approximately 2.21 × 10-6 M.
To determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 4.91, we need to use the Ksp (solubility product constant) of Zn(CN)2 and the dissociation reaction of HCN (hydrogen cyanide) in water.
Step 1: Write the balanced equation for the dissociation of Zn(CN)2 in water.
Zn(CN)2(s) ⇌ Zn2+(aq) + 2CN-(aq)
The equilibrium expression for this reaction is:
Ksp = [Zn2+][CN-]^2
Step 2: Write the balanced equation for the dissociation of HCN in water.
HCN(aq) ⇌ H+(aq) + CN-(aq)
The equilibrium expression for this reaction is:
Ka = [H+][CN-]/[HCN]
Step 3: Identify the relationship between HCN and CN-.
HCN and CN- share the same concentration when they are in equilibrium.
Step 4: Determine the concentration of CN-.
Since HCN is a weak acid, we can use the given pH and the Ka value to find the concentration of HCN, and since HCN and CN- share the same concentration, we can use the concentration of HCN to determine the concentration of CN-.
Using the formula for pH:
pH = -log[H+]
[H+] = 10^(-pH)
[H+] = 10^(-4.91) = 7.13 × 10^(-5) M
Using the Ka expression:
Ka = [H+][CN-]/[HCN]
[H+] = [CN-] (Let's call this concentration x)
[HCN] = initial concentration of HCN (which is equal to its initial molar solubility, S, since it is a weak acid)
Ka = x^2/S
Since the concentration of CN- is x and the concentration of HCN is S, we can rewrite the equation as:
6.2 × 10^(-10) = x^2/S
Step 5: Calculate the molar solubility (S) of Zn(CN)2 using the Ksp value.
Now, substitute the value of [CN-] (which is x) into the Ksp expression:
Ksp = [Zn2+][CN-]^2
3.0 × 10^(-16) = [Zn2+](x)^2
Step 6: Solve the system of equations.
We have two equations:
6.2 × 10^(-10) = x^2/S
3.0 × 10^(-16) = [Zn2+](x)^2
Solve these equations simultaneously to find the values of x and S.
Once you solve the system, you will find the concentration of CN- (x) and the molar solubility (S) of Zn(CN)2 in the given solution with pH = 4.91.
pH 4.91; therefore, (H^+) = 1.23E-5M
This is worked like your other solubility problem.
Zn(CN)2 ==> Zn^+ + 2CN^-
2H^+ + 2CN^- ==> 2HCN add the two equns
------------------------------------
Zn(CN)2 + 2H^+ ==> Zn^2+ + 2HCN
.........1.23E-5....x.......2x
Keq = Ksp/Ka = 3.0E-16/(6.2E-10)^2 = about 780 but you need to confirm that.
780 = (x)(x)/(1.23E-5)^2 and solve for x.