`K_(a)` for `HCN` is `5xx10^(-10)` at `25^(@)C`. For maintaining a constant pH of 9, the volume in ml of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution is

Answer

To solve the problem, we need to determine the volume of a 5 M KCN solution required to maintain a pH of 9 when added to 10 ml of a 2 M HCN solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the pKa of HCN
Given the dissociation constant \( K_a \) for HCN is \( 5 \times 10^{-10} \).

Using the formula:
\[ pK_a = -\log(K_a) \]
\[ pK_a = -\log(5 \times 10^{-10}) \approx 9.30 \]

### Step 2: Set up the Henderson-Hasselbalch equation
To maintain a pH of 9, we can use the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \( [A^-] \) is the concentration of the salt (KCN)
- \( [HA] \) is the concentration of the weak acid (HCN)

Substituting the known values:
\[ 9 = 9.30 + \log\left(\frac{[KCN]}{[HCN]}\right) \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \log\left(\frac{[KCN]}{[HCN]}\right) = 9 - 9.30 = -0.30 \]
This implies:
\[ \frac{[KCN]}{[HCN]} = 10^{-0.30} \approx 0.50 \]

### Step 4: Calculate the concentration of HCN
The concentration of HCN in the solution can be calculated as follows:
- Volume of HCN solution = 10 ml
- Concentration of HCN = 2 M

Calculating moles of HCN:
\[ \text{Moles of HCN} = \text{Concentration} \times \text{Volume} = 2 \, \text{mol/L} \times 0.010 \, \text{L} = 0.02 \, \text{mol} = 20 \, \text{mmol} \]

### Step 5: Calculate the moles of KCN required
Using the ratio from the previous step:
\[ \frac{[KCN]}{[HCN]} = 0.50 \implies [KCN] = 0.50 \times [HCN] \]
Let \( n \) be the moles of KCN. Then:
\[ n = 0.50 \times 20 \, \text{mmol} = 10 \, \text{mmol} \]

### Step 6: Calculate the volume of KCN solution needed
The concentration of KCN is 5 M, which means:
\[ \text{Moles} = \text{Concentration} \times \text{Volume} \]
Rearranging gives:
\[ \text{Volume} = \frac{\text{Moles}}{\text{Concentration}} = \frac{10 \, \text{mmol}}{5 \, \text{mol/L}} = \frac{10 \times 10^{-3} \, \text{mol}}{5 \, \text{mol/L}} = 2 \, \text{ml} \]

### Conclusion
Thus, the volume of 5 M KCN solution required to be added to 10 ml of 2 M HCN solution to maintain a pH of 9 is **2 ml**.

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