A solution of `0.1M NaZ` has `pH = 8.90`. The `K_(a)` of `HZ` is

Options

A. `6.3 xx 10^(-11)`
B. `6.3 xx 10^(-10)`
C. `1.6 xx 10^(-5)`
D. `1.6 xx 10^(-6)`

Explanation

To find the \( K_a \) of \( HZ \) from the given information about the solution of \( 0.1M \, NaZ \) with a pH of 8.90, we can follow these steps:

### Step 1: Understand the Hydrolysis of the Salt
The salt \( NaZ \) is formed from a strong base (\( NaOH \)) and a weak acid (\( HZ \)). When \( NaZ \) dissolves in water, it hydrolyzes to produce \( OH^- \) ions and \( HZ \):
\[ NaZ \, \text{(s)} \rightarrow Na^+ \, \text{(aq)} + Z^- \, \text{(aq)} \]
\[ Z^- + H_2O \rightleftharpoons HZ + OH^- \]

### Step 2: Determine the pOH
Since the pH of the solution is given as 8.90, we can calculate the pOH:
\[ pOH = 14 - pH = 14 - 8.90 = 5.10 \]

### Step 3: Calculate the Concentration of Hydroxide Ions
Using the pOH, we can find the concentration of hydroxide ions \( [OH^-] \):
\[ [OH^-] = 10^{-pOH} = 10^{-5.10} \approx 7.94 \times 10^{-6} \, M \]

### Step 4: Set Up the Equilibrium Expression
The hydrolysis reaction can be represented by the equilibrium constant \( K_b \):
\[ K_b = \frac{[HZ][OH^-]}{[Z^-]} \]
Assuming that the initial concentration of \( Z^- \) is \( 0.1 \, M \) and \( x \) is the amount that reacts, at equilibrium:
\[ [HZ] = x, \quad [OH^-] = x, \quad [Z^-] = 0.1 - x \approx 0.1 \, M \, (since \, x \, is \, very \, small) \]

### Step 5: Substitute into the Equilibrium Expression
Substituting the values into the \( K_b \) expression:
\[ K_b = \frac{x \cdot x}{0.1} = \frac{x^2}{0.1} \]
Since \( x = [OH^-] \approx 7.94 \times 10^{-6} \):
\[ K_b = \frac{(7.94 \times 10^{-6})^2}{0.1} \approx \frac{6.30 \times 10^{-11}}{0.1} = 6.30 \times 10^{-10} \]

### Step 6: Relate \( K_a \) and \( K_b \)
Using the relation \( K_a \cdot K_b = K_w \) where \( K_w = 1.0 \times 10^{-14} \):
\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{6.30 \times 10^{-10}} \approx 1.58 \times 10^{-5} \]

### Step 7: Final Answer
Thus, the \( K_a \) of \( HZ \) is approximately:
\[ K_a \approx 1.6 \times 10^{-5} \]