What is the percentage hydrolysis of NaCN is `N//80` soluiton, when the dissociation constant for HCN is `1.3 xx 10^(-9)` and `K_(w) =1.0 xx 10^(-14)`

Options

A. 2.48
B. 5.26
C. 8.2
D. 9.6

Explanation

To find the percentage hydrolysis of NaCN in an N/80 solution, we can follow these steps:

### Step 1: Understand the given data
- **Dissociation constant for HCN (Ka)** = \(1.3 \times 10^{-9}\)
- **Ion product of water (Kw)** = \(1.0 \times 10^{-14}\)
- **Concentration of NaCN** = \(N/80\) = \(1/80\) N

### Step 2: Calculate the hydrolysis constant (Kh)
The hydrolysis constant \(Kh\) can be calculated using the relationship:
\[ Kh = \frac{Kw}{Ka} \]

Substituting the values:
\[ Kh = \frac{1.0 \times 10^{-14}}{1.3 \times 10^{-9}} = \frac{1.0}{1.3} \times 10^{-5} = 0.7692 \times 10^{-5} = 7.692 \times 10^{-6} \]

### Step 3: Calculate the concentration of NaCN in molarity
Since the concentration is given in normality, and for NaCN (which is a monobasic salt), the normality is equal to the molarity:
\[ C = \frac{1}{80} \text{ N} = \frac{1}{80} \text{ M} \]

### Step 4: Calculate the degree of hydrolysis (h)
Using the formula for degree of hydrolysis \(h\):
\[ h = \sqrt{\frac{Kh}{C}} \]
Substituting the values:
\[ h = \sqrt{\frac{7.692 \times 10^{-6}}{\frac{1}{80}}} \]
\[ h = \sqrt{7.692 \times 10^{-6} \times 80} = \sqrt{6.1536 \times 10^{-4}} \approx 0.0248 \]

### Step 5: Calculate the percentage hydrolysis
To find the percentage hydrolysis, we multiply the degree of hydrolysis by 100:
\[ \text{Percentage Hydrolysis} = h \times 100 = 0.0248 \times 100 = 2.48\% \]

### Final Answer
The percentage hydrolysis of NaCN in an N/80 solution is approximately **2.48%**.

---