`K_a` for cyanoacetic acid is `4 xx 10^(-3_`. What is the value of degree of hydrolysis of 0.4 M sodium cy- ano acetate solution ?

Options

A. `4.5 xx 10^(-6)`
B. `5.5 xx 10^(-6)`
C. `2.5 xx 10^(-6)`
D. `3.5 xx 10^(-6)`

Explanation

To find the degree of hydrolysis of a 0.4 M sodium cyanoacetate solution, we will follow these steps:

### Step 1: Understand the System
We have cyanoacetic acid (weak acid) and sodium cyanoacetate (its salt, which provides the conjugate base). The salt will undergo hydrolysis in water, producing hydroxide ions (OH⁻) and the weak acid (HCN).

### Step 2: Write the Hydrolysis Reaction
The hydrolysis of the cyanoacetate ion (CN⁻) can be represented as:
\[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \]

### Step 3: Write the Expression for \(K_h\)
The equilibrium constant for this hydrolysis reaction, \(K_h\), can be expressed as:
\[ K_h = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \]

### Step 4: Relate \(K_h\), \(K_a\), and \(K_w\)
Using the relationship between \(K_h\), \(K_a\) (for cyanoacetic acid), and \(K_w\) (the ion product of water):
\[ K_h = \frac{K_w}{K_a} \]
Where:
- \(K_w = 1.0 \times 10^{-14}\) at 25°C
- \(K_a = 4.0 \times 10^{-3}\)

### Step 5: Calculate \(K_h\)
Substituting the values into the equation:
\[ K_h = \frac{1.0 \times 10^{-14}}{4.0 \times 10^{-3}} = 2.5 \times 10^{-12} \]

### Step 6: Set Up the Hydrolysis Concentration Equation
Let \(H\) be the degree of hydrolysis. The concentration of CN⁻ at equilibrium will be:
\[ [\text{CN}^-] = 0.4 - H \]
The concentrations of HCN and OH⁻ at equilibrium will both be \(H\).

### Step 7: Substitute into the \(K_h\) Expression
Substituting the equilibrium concentrations into the \(K_h\) expression:
\[ K_h = \frac{H \cdot H}{0.4 - H} = \frac{H^2}{0.4 - H} \]

### Step 8: Assume \(H\) is Small
Assuming \(H\) is small compared to 0.4 M, we can simplify:
\[ K_h \approx \frac{H^2}{0.4} \]

### Step 9: Solve for \(H\)
Now substituting the value of \(K_h\):
\[ 2.5 \times 10^{-12} = \frac{H^2}{0.4} \]
\[ H^2 = 2.5 \times 10^{-12} \times 0.4 \]
\[ H^2 = 1.0 \times 10^{-12} \]
\[ H = \sqrt{1.0 \times 10^{-12}} = 1.0 \times 10^{-6} \]

### Step 10: Conclusion
Thus, the degree of hydrolysis \(H\) of the sodium cyanoacetate solution is:
\[ H = 1.0 \times 10^{-6} \]