pOH of 0.1 molar aqueous NaCN solution found to be 2, then calculate value of dissociation constant of HCN in its aqueous solution.

Options

A. `1.11xx10^(-3)`
B. `10^(-2)`
C. `10^(-11)`
D. `9xx10^(-12)`

Answer

To calculate the dissociation constant of HCN (Ka) from the given information about a 0.1 molar aqueous NaCN solution with a pOH of 2, we can follow these steps:

### Step 1: Convert pOH to [OH⁻]
Given that pOH = 2, we can find the concentration of hydroxide ions ([OH⁻]) using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Thus,
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{mol/L} \]

### Step 2: Set up the hydrolysis reaction
NaCN is a salt derived from a strong base (NaOH) and a weak acid (HCN). In solution, CN⁻ will hydrolyze according to the reaction:
\[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \]
The initial concentration of CN⁻ is 0.1 M.

### Step 3: Establish equilibrium concentrations
Let \( H \) be the degree of hydrolysis. At equilibrium:
- Concentration of CN⁻ = \( 0.1(1 - H) \)
- Concentration of HCN = \( 0.1H \)
- Concentration of OH⁻ = \( 0.1H \)

From Step 1, we know:
\[ [\text{OH}^-] = 10^{-2} \, \text{mol/L} = 0.1H \]
Thus,
\[ H = \frac{10^{-2}}{0.1} = 10^{-1} = 0.1 \]

### Step 4: Calculate the hydrolysis constant (Kh)
The hydrolysis constant (Kh) can be expressed as:
\[ K_H = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \]
Substituting the equilibrium concentrations:
\[ K_H = \frac{(0.1)(0.1)}{0.1(1 - 0.1)} = \frac{0.01}{0.1 \times 0.9} = \frac{0.01}{0.09} = \frac{10^{-2}}{0.09} \]

### Step 5: Relate Kh to Ka
For the dissociation of HCN:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]
The dissociation constant (Ka) is given by:
\[ K_a = \frac{[\text{H}^+][\text{CN}^-]}{[\text{HCN}]} \]
Using the ionic product of water (Kw):
\[ K_w = [\text{H}^+][\text{OH}^-] = 10^{-14} \]
From Step 1, we have:
\[ [\text{H}^+] = \frac{10^{-14}}{10^{-2}} = 10^{-12} \]

### Step 6: Substitute values into Ka expression
At equilibrium:
- [CN⁻] = \( 0.1(1 - 0.1) = 0.1 \times 0.9 = 0.09 \)
- [HCN] = \( 0.1H = 0.1 \times 0.1 = 0.01 \)

Thus,
\[ K_a = \frac{(10^{-12})(0.09)}{0.01} = \frac{9 \times 10^{-12}}{0.01} = 9 \times 10^{-12} \]

### Final Answer
The dissociation constant \( K_a \) of HCN is:
\[ K_a = 9 \times 10^{-12} \]