Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 1.94. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.
See your Ag2CO3 post.
To determine the molar solubility (S) of Zn(CN)2 in a solution with a pH of 1.94, we need to consider the equilibrium between Zn(CN)2 and its ions in solution.
The balanced equation for the dissolution of Zn(CN)2 in water is:
Zn(CN)2(s) ⇌ Zn2+(aq) + 2CN-(aq)
Given that the Ksp (solubility product constant) for Zn(CN)2 is 3.0 × 10^(-16), we know that the equilibrium expression for this reaction is as follows:
Ksp = [Zn2+][CN-]^2
Since the concentration of water (H2O) remains constant, we can ignore it in the equilibrium expression. Therefore, the equation becomes:
Ksp = [Zn2+][CN-]^2
Next, we need to consider how the pH of the solution affects the concentrations of Zn2+ and CN- ions.
The pH of a solution is a measure of the concentration of hydrogen ions (H+). In this case, a pH value of 1.94 indicates a high concentration of H+ ions in the solution. Since the pH is below the pKa of HCN (which is approximately equal to -log10(Ka)), we can assume that HCN is mostly in its undissociated form.
Therefore, we can write an equilibrium equation for the dissociation of HCN as follows:
HCN(aq) ⇌ H+(aq) + CN-(aq)
The equilibrium expression for this dissociation reaction is:
Ka = [H+][CN-]/[HCN]
Given that the Ka for HCN is 6.2 × 10^(-10), we can rearrange the equation to solve for [CN-]:
[Cn-] = Ka * [HCN] / [H+]
Substituting [H+] with 10^(-pH) (since pH is equal to -log10([H+])), the equation becomes:
[CN-] = Ka * [HCN] / 10^(-pH)
Since the concentration of HCN is equal to the molar solubility (S) of Zn(CN)2 (since we assume that all Zn(CN)2 dissociates into Zn2+ and CN- ions), we can rewrite the equation as:
[CN-] = Ka * S / 10^(-pH)
Now we have an equation that relates the concentration of CN- to the molar solubility (S) of Zn(CN)2 and the pH of the solution.
To find the molar solubility (S), we need to solve for [CN-]. Rearranging the equation, we get:
S = [CN-] * 10^(-pH) / Ka
Substituting the values we have:
S = [CN-] * 10^(-1.94) / (6.2 × 10^(-10))
Since the concentration of CN- is twice the concentration of Zn2+ (according to the balanced equation), and we assume that the molar solubility (S) is equal to the concentration of Zn(CN)2, we can write:
[CN-] = 2S
Finally, substituting this value, we have:
S = 2S * 10^(-1.94) / (6.2 × 10^(-10))
Solving this equation will give you the molar solubility (S) of Zn(CN)2 in a solution with a pH of 1.94.