The pH of a solution of `NH_4Cl` is 5.86. calculate the molar concentration of the solution if Kb =1.0×`10^(−5)` and Kw=1.0×`10^(−14)` .

Options

A. 0.19M
B. 0.019M
C. 1.9M
D. 1.09M

Explanation

To solve the problem, we need to calculate the molar concentration of the solution of \( NH_4Cl \) given its pH and the values of \( K_b \) and \( K_w \).

### Step-by-Step Solution:

1. **Identify the given values:**
- pH of the solution = 5.86
- \( K_b = 1.0 \times 10^{-5} \)
- \( K_w = 1.0 \times 10^{-14} \)

2. **Calculate \( pK_w \) and \( pK_b \):**
- \( pK_w = -\log(K_w) = -\log(1.0 \times 10^{-14}) = 14 \)
- \( pK_b = -\log(K_b) = -\log(1.0 \times 10^{-5}) = 5 \)

3. **Calculate \( pK_a \) using the relationship:**
\[ pK_a + pK_b = pK_w \]
\[ pK_a = pK_w - pK_b = 14 - 5 = 9 \]

4. **Use the pH to find the concentration of \( H^+ \):**
\[ [H^+] = 10^{-pH} = 10^{-5.86} \approx 1.38 \times 10^{-6} \, M \]

5. **Use the relationship between \( K_a \), \( [H^+] \), and concentration \( C \) of \( NH_4^+ \):**
For the dissociation of \( NH_4^+ \):
\[ NH_4^+ \rightleftharpoons H^+ + NH_3 \]
The expression for \( K_a \) is:
\[ K_a = \frac{[H^+][NH_3]}{[NH_4^+]} \]
Assuming \( [NH_3] \approx [H^+] \) and \( [NH_4^+] \approx C - [H^+] \approx C \) (since \( [H^+] \) is very small compared to \( C \)):
\[ K_a = \frac{(1.38 \times 10^{-6})(1.38 \times 10^{-6})}{C} \]

6. **Substituting \( K_a \) into the equation:**
\[ K_a = 10^{-pK_a} = 10^{-9} \]
Therefore,
\[ 10^{-9} = \frac{(1.38 \times 10^{-6})^2}{C} \]

7. **Rearranging to find \( C \):**
\[ C = \frac{(1.38 \times 10^{-6})^2}{10^{-9}} = \frac{1.9044 \times 10^{-12}}{10^{-9}} = 1.9044 \times 10^{-3} \, M \]

8. **Final result:**
\[ C \approx 0.0019 \, M \]

### Conclusion:
The molar concentration of the solution of \( NH_4Cl \) is approximately \( 0.0019 \, M \).