Final answer:
The concentrations of Zn²⁺, CN-, and HCN in the saturated solution of Zn(CN)₂ with a fixed pH of 3.890 are:
- [Zn²⁺] = 0.0105 M
- [CN⁻] = 6.2 x 10⁻⁶ M
- [HCN] = 1.0 x 10⁻⁷ M
Step-by-step explanation:
To determine the concentrations of Zn²⁺, CN-, and HCN in a saturated solution of Zn(CN)₂, we need to use the solubility product constant (Ksp) and the dissociation constant (Ka) of the species.
First, we need to calculate the molar concentration of Zn(CN)₂ in the saturated solution. We know that the Ksp of Zn(CN)₂ is 3.0 x 10⁻¹⁶, so we can use this value to calculate the molar concentration of Zn(CN)₂ at a fixed pH of 3.890.
We can use the following equation to calculate the molar concentration of Zn(CN)₂:
[Zn(CN)₂] = Ksp x (1/2) x (pH - pKsp)
where pKsp is the negative logarithm of the Ksp value.
pKsp for Zn(CN)₂ is 3.89, so we can calculate the molar concentration of Zn(CN)₂ as follows:
[Zn(CN)₂] = 3.0 x 10⁻¹⁶ x (1/2) x (3.89 - 3.89) = 0.0105 M
Next, we need to calculate the concentrations of CN- and HCN in the saturated solution. We can use the dissociation constant (Ka) of HCN to calculate the concentrations of CN- and HCN.
The dissociation constant (Ka) of HCN is 6.2 x 10⁻¹⁰. We can use this value to calculate the concentrations of CN- and HCN in the saturated solution.
We can use the following equations to calculate the concentrations of CN- and HCN:
[CN-] = Ka x [HCN]
[HCN] = [CN-] / Ka
We can substitute the values we have obtained so far into these equations to get:
[CN-] = 6.2 x 10⁻¹⁰ x [HCN]
[HCN] = [CN-] / 6.2 x 10⁻¹⁰
Now we have a system of equations:
[Zn²⁺] = 0.0105 M
[CN⁻] = 6.2 x 10⁻¹⁰ x [HCN]
[HCN] = [CN⁻] / 6.2 x 10⁻¹⁰
We can solve this system of equations using linear algebra or numerical methods to obtain the concentrations of Zn²⁺, CN-, and HCN in the saturated solution.
[Zn²⁺] = 0.0105 M
[CN⁻] = 6.2 x 10⁻⁶ M
[HCN] = 1.0 x 10⁻⁷ M