To find the concentration of \([CN^-]\) in a solution prepared by mixing 100 mL of 0.1 M KCN and 100 mL of 0.1 M HCl, we can follow these steps:
### Step 1: Determine the initial moles of KCN and HCl
- **KCN**:
\[
\text{Moles of KCN} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles}
\]
- **HCl**:
\[
\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.1 \, \text{M} \times 0.1 \, \text{L} = 0.01 \, \text{moles}
\]
### Step 2: Identify the reaction
When KCN is mixed with HCl, the following reaction occurs:
\[
KCN + HCl \rightarrow HCN + KCl
\]
This means that KCN will react with HCl to produce HCN and KCl.
### Step 3: Determine the limiting reagent
Since both KCN and HCl are present in equal moles (0.01 moles each), they will completely react with each other. Therefore, neither is in excess.
### Step 4: Calculate the moles of HCN produced
After the reaction:
- Moles of HCN produced = Moles of KCN reacted = 0.01 moles
### Step 5: Calculate the concentration of HCN in the mixed solution
The total volume of the solution after mixing is:
\[
\text{Total Volume} = 100 \, \text{mL} + 100 \, \text{mL} = 200 \, \text{mL} = 0.2 \, \text{L}
\]
The concentration of HCN is:
\[
\text{Concentration of HCN} = \frac{\text{Moles of HCN}}{\text{Total Volume}} = \frac{0.01 \, \text{moles}}{0.2 \, \text{L}} = 0.05 \, \text{M}
\]
### Step 6: Use the dissociation constant to find \([CN^-]\)
The dissociation of HCN can be represented as:
\[
HCN \rightleftharpoons H^+ + CN^-
\]
The \(K_a\) expression is given by:
\[
K_a = \frac{[H^+][CN^-]}{[HCN]}
\]
Given \(K_a = 5 \times 10^{-4}\) and assuming that \(x\) is the concentration of \(CN^-\) produced, we can set up the equation:
\[
K_a = \frac{(x)(x)}{0.05 - x} \approx \frac{x^2}{0.05}
\]
Assuming \(x\) is small compared to 0.05, we can simplify:
\[
5 \times 10^{-4} = \frac{x^2}{0.05}
\]
\[
x^2 = 5 \times 10^{-4} \times 0.05
\]
\[
x^2 = 2.5 \times 10^{-5}
\]
\[
x = \sqrt{2.5 \times 10^{-5}} \approx 5 \times 10^{-3} \, \text{M}
\]
### Step 7: Calculate the final concentration of \([CN^-]\)
Since \(x\) represents the concentration of \(CN^-\):
\[
[CN^-] \approx 5 \times 10^{-3} \, \text{M}
\]
### Final Result
The concentration of \([CN^-]\) in the solution is approximately \(5 \times 10^{-3} \, \text{M}\).
---
