Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 1.78. The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

probably because your explanation is very hard to follow. Have seen three of these answered by you and couldn't make heads or tails of any of it.

bad

Determine [Zn2 ], [CN–], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 2.390 The Ksp for Zn(CN)2 is 3.0 × 10–16. The Ka for HCN is 6.2 × 10–10.

To determine the concentrations of [Zn2+], [CN-], and [HCN] in a saturated solution of Zn(CN)2 with a fixed pH of 1.78, we need to use the given Ksp for Zn(CN)2 and Ka for HCN.

Let's work through the problem step by step:

1. Start by writing the balanced equation for the dissociation of Zn(CN)2 in water:

Zn(CN)2 ⇌ Zn2+ + 2CN-

From this equation, we can see that for every 1 mole of Zn(CN)2 that dissociates, we get 1 mole of Zn2+ and 2 moles of CN-.

2. Since the solution is saturated, it means that Zn(CN)2 is at its maximum solubility and is in equilibrium with its ions. Therefore, we assume that the concentrations of Zn2+ and CN- are equal and represent them as 'x'.

[Zn2+] = [CN-] = x

3. Next, use the given Ksp value of 3.0 × 10–16 to set up the solubility product expression:

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

Substituting the values, we get:
3.0 × 10–16 = (x)(x^2)

4. Solve the equation to find 'x', which represents both [Zn2+] and [CN-]. In this case, we have a quadratic equation:

x^3 = 3.0 × 10–16

Taking the cube root of both sides, we find:
x ≈ 6.1 × 10–6

Therefore, [Zn2+] ≈ [CN-] ≈ 6.1 × 10–6 M.

5. Lastly, let's determine the concentration of [HCN] in the saturated solution. We use the given Ka value of 6.2 × 10–10 to set up the acid dissociation equation:

HCN ⇌ H+ + CN-

From this equation, we can see that for every 1 mole of HCN that dissociates, we get 1 mole of H+ and 1 mole of CN-.

Since the pH of the solution is given as 1.78, we know the concentration of [H+] is 10^(-pH). Calculate it:

[H+] = 10^(-1.78) ≈ 1.59 × 10^(-2) M

Since the concentration of [CN-] is approximately 6.1 × 10^(-6) M (as determined previously), and the dissociation of HCN is 1:1, the concentration of [HCN] will be approximately the same as [CN-].

Therefore, [HCN] ≈ 6.1 × 10^(-6) M.

To summarize:
- [Zn2+] ≈ [CN-] ≈ 6.1 × 10^(-6) M
- [HCN] ≈ 6.1 × 10^(-6) M

I worked this for you a night or so ago. Why do you need it again?