To solve the problem, we need to find the concentration of the cyanide ion \([CN^-]\) in a solution prepared by mixing 100 mL of 0.1 M KCN and 100 mL of 0.1 M HCl. We also know the dissociation constant \(K_a\) of HCN is \(5 \times 10^{-6}\).
### Step-by-Step Solution:
1. **Calculate the Initial Molarities After Mixing:**
When we mix equal volumes of KCN and HCl, the total volume becomes:
\[
V_{\text{total}} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL}
\]
The initial concentrations of KCN and HCl before any reaction occurs are:
\[
[KCN] = \frac{0.1 \, \text{mol/L} \times 100 \, \text{mL}}{200 \, \text{mL}} = 0.05 \, \text{M}
\]
\[
[HCl] = \frac{0.1 \, \text{mol/L} \times 100 \, \text{mL}}{200 \, \text{mL}} = 0.05 \, \text{M}
\]
2. **Set Up the Equilibrium Expression:**
The dissociation of HCN can be represented as:
\[
HCN \rightleftharpoons H^+ + CN^-
\]
The equilibrium expression for this reaction is given by:
\[
K_a = \frac{[H^+][CN^-]}{[HCN]}
\]
Given \(K_a = 5 \times 10^{-6}\).
3. **Determine the Concentrations at Equilibrium:**
Let \(x\) be the amount of HCN that dissociates. At equilibrium, we have:
\[
[H^+] = 0.05 + x \quad \text{(since HCl is a strong acid, it completely dissociates)}
\]
\[
[CN^-] = x
\]
\[
[HCN] = 0.05 - x
\]
4. **Substitute into the Equilibrium Expression:**
Substituting these values into the \(K_a\) expression gives:
\[
5 \times 10^{-6} = \frac{(0.05 + x)(x)}{(0.05 - x)}
\]
Since \(x\) is small compared to 0.05, we can approximate:
\[
5 \times 10^{-6} \approx \frac{(0.05)(x)}{0.05} = x
\]
Thus, we find:
\[
x = 5 \times 10^{-6}
\]
5. **Calculate the Concentration of \([CN^-]\):**
Therefore, the concentration of cyanide ions \([CN^-]\) in the solution is:
\[
[CN^-] = x = 5 \times 10^{-6} \, \text{M}
\]
### Final Answer:
The concentration of \([CN^-]\) in the solution is \(5 \times 10^{-6} \, \text{M}\).
