Question

$20 \,mL$ of $01 \,M \,NH _4 OH$ is mixed with $40\, mL$ of $005\, M \,HCl$ The $pH$ of the mixture is nearest to: (Given: $K _{ b }\left( NH _4 OH \right)=1 \times 10^{-5}, \log 2=030$, $\log 3=048, \log 5=069, \log 7=084$, $\log 11=104$ )

• A. $3.2$
• B. $4.2$
• C. $5.2$
• D. $6.2$

Answer 1

The Correct Option is C

To solve this problem, we need to determine the pH of the mixture of $20 \, \text{mL}$ of $0.1 \, \text{M} \, \text{NH}_4\text{OH}$ and $40 \, \text{mL}$ of $0.05 \, \text{M} \, \text{HCl}$. The given ammonia solution ($\text{NH}_4\text{OH}$) acts as a weak base, and the hydrochloric acid ($\text{HCl}$) is a strong acid.

1. First, calculate the initial moles of $\text{NH}_4\text{OH}$ and $\text{HCl}$.
   • Moles of $\text{NH}_4\text{OH}$: \(0.1 \times \frac{20}{1000} = 0.002 \, \text{moles}\)
   • Moles of $\text{HCl}$: \(0.05 \times \frac{40}{1000} = 0.002 \, \text{moles}\)
2. Since the moles of $\text{NH}_4\text{OH}$ are equal to moles of $\text{HCl}$, they neutralize each other completely, forming $\text{NH}_4\text{Cl}$.

The reaction is:

1. \(\text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O}\)
2. The resulting solution contains $0.002$ moles of $\text{NH}_4\text{Cl}$ in a total volume of $60 \, \text{mL}$ (since $20 \, \text{mL} + 40 \, \text{mL} = 60 \, \text{mL}$). Now, we need to calculate the concentration of $\text{NH}_4\text{Cl}$.
3. Concentration of $\text{NH}_4\text{Cl}$:
   • \(C = \frac{0.002}{0.060} = \frac{1}{30} = 0.0333 \, \text{M}\)
4. The $\text{NH}_4^+$ ion in solution will hydrolyze to form $\text{NH}_3$ and $\text{H}^+$.

The hydrolysis reaction is:

1. \(\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+\)
2. For $\text{NH}_4^+$:
   • \(K_w = K_a \cdot K_b\)
   • \(K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-5}} = 10^{-9}\)
3. The expression for hydrolysis constant \(K_h\) is:
   • \(K_h = \frac{K_w}{K_b}\)
4. Using the relation between \(pH\) and \(pK_a\):
   • \([\text{H}^+] = \sqrt{C \cdot K_a}\)
   • \([\text{H}^+] = \sqrt{0.0333 \cdot 10^{-9}} = 5.77 \times 10^{-6} \, \text{M}\)
   • Calculate the pH: \(\text{pH} = -\log(5.77 \times 10^{-6})\)
   • \(\text{pH} \approx 5.2\)

Thus, the pH of the mixture is approximately 5.2. This matches the given correct answer option.