Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 3.63. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.

I'm completely lost help please!

You can work this one of two ways. Here is one way; I don't know which would be easier for you to understand.

Zn(CN)2 ==> Zn + 2CN^-
Then 2CN^- + 2H^+ ==> 2HCN.
Solubility Zn(CN)2 = S
Then Zn^2+ = S and
CN^- = 2S. The total CN = 2S and
2S = (CN^-) + (HCN)

Use Ka for HCN = (H^+)(CN^-)/(HCN)
Solve for (HCN)/(CN^-) = and convert that to (HCN) = (CN^-)*some number.
Then go back to 2S = CN + HCN
and this becomes 2S = CN + somenumber*CN
and convert this to CN = ? in terms of S.
Plug this back into Ksp and solve for S. I get a value of 0.022 M. Try this and see if you understand. If so there is no need to go through the other method.

this doesn't make sense

To determine the molar solubility (S) of Zn(CN)2 in a solution with a pH of 3.63, we need to consider the equilibrium of the dissociation of Zn(CN)2 in water.

The first step is to write down the balanced equation for the dissociation of Zn(CN)2:

Zn(CN)2 ⇌ Zn2+ + 2CN-

Next, we need to look at the relevant equilibrium constant, which is the solubility product constant (Ksp) for Zn(CN)2. The Ksp expression for the dissociation of Zn(CN)2 is:

Ksp = [Zn2+][CN-]^2

Given that the Ksp value for Zn(CN)2 is 3.0 × 10^-16, we can use this information to determine the concentrations of the ions at equilibrium.

Let's assume that the molar solubility of Zn(CN)2 is S mol/L. Therefore, at equilibrium, the concentration of Zn2+ will be S mol/L, and the concentration of CN- will be 2S mol/L.

To proceed, we need to make use of the pH value in the question. The pH of a solution is a measure of its acidity and is related to the concentration of hydrogen ions (H+) present. In this case, we are given that the pH is 3.63, which means the concentration of H+ is 10^(-pH) = 10^(-3.63) mol/L.

Now, we need to consider the dissociation of hydrocyanic acid (HCN) in water:

HCN ⇌ H+ + CN-

The equilibrium constant for this reaction is the acid dissociation constant (Ka), which is given as 6.2 × 10^-10. Using the information provided, we can set up an expression for Ka:

Ka = [H+][CN-]/[HCN]

At equilibrium, the concentration of CN- is 2S mol/L. Therefore, the concentration of HCN is (10^(-3.63) - [H+]) mol/L.

Substituting these values into the Ka expression, we get:

6.2 × 10^-10 = [H+](2S)/(10^(-3.63) - [H+])

Now, we have an equation with one unknown, [H+]. We can solve this equation to find the concentration of H+ at equilibrium. Once we have the value for [H+], we can substitute it back into the equation for Ka to determine the value of S, the molar solubility of Zn(CN)2.

It is important to note that this calculation assumes ideal behavior and neglects any potential complex formations or other factors that may impact the solubility of Zn(CN)2.