The pH of a solution of `NH_4Cl` is 4.86 .Calculate the molar concentration fo the solution if ` K_b= 1.0 xx 10^(-5) and K_w = 1.0 xx 10 ^(-14) `

Answer

To solve the problem, we need to calculate the molar concentration of the ammonium chloride (NH₄Cl) solution given its pH, \( K_b \), and \( K_w \).

### Step-by-Step Solution:

1. **Determine pH and pOH**:
Given that the pH of the solution is 4.86, we can calculate the pOH using the relationship:
\[ pH + pOH = 14 \]
Thus,
\[ pOH = 14 - pH = 14 - 4.86 = 9.14 \]

2. **Calculate \( K_a \)**:
Since \( NH_4Cl \) is a salt of a weak base (NH₄OH) and a strong acid (HCl), we can find the \( K_a \) of the ammonium ion (NH₄⁺) using the relationship:
\[ K_a \cdot K_b = K_w \]
Rearranging gives:
\[ K_a = \frac{K_w}{K_b} \]
Substituting the values:
\[ K_a = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9} \]

3. **Set up the equilibrium expression**:
The dissociation of \( NH_4^+ \) in water can be represented as:
\[ NH_4^+ \rightleftharpoons NH_3 + H^+ \]
The equilibrium expression for this reaction is:
\[ K_a = \frac{[NH_3][H^+]}{[NH_4^+]} \]
Let the concentration of \( NH_4^+ \) be \( C \) (the concentration we want to find). At equilibrium, the concentration of \( H^+ \) will be equal to \( x \) (which can be derived from the pH), and the concentration of \( NH_3 \) will also be \( x \).

4. **Calculate \( [H^+] \)**:
From the pH, we can find \( [H^+] \):
\[ [H^+] = 10^{-pH} = 10^{-4.86} \approx 1.38 \times 10^{-5} \, \text{M} \]

5. **Substituting into the equilibrium expression**:
Now substituting into the equilibrium expression:
\[ K_a = \frac{(1.38 \times 10^{-5})(1.38 \times 10^{-5})}{C - 1.38 \times 10^{-5}} \approx \frac{(1.38 \times 10^{-5})^2}{C} \]
Since \( C \) is much larger than \( 1.38 \times 10^{-5} \), we can approximate \( C - 1.38 \times 10^{-5} \approx C \).

6. **Solve for \( C \)**:
Rearranging gives:
\[ C = \frac{(1.38 \times 10^{-5})^2}{K_a} = \frac{(1.38 \times 10^{-5})^2}{1.0 \times 10^{-9}} \]
Calculating this gives:
\[ C = \frac{1.9044 \times 10^{-10}}{1.0 \times 10^{-9}} = 0.1904 \, \text{M} \]

7. **Final Answer**:
Therefore, the molar concentration of the \( NH_4Cl \) solution is approximately:
\[ C \approx 0.19 \, \text{M} \]