Question

Degree of dissociation of an acid HCl is $95%$, $0.192\mathrm{g}$ of the acid is present in $0.5\mathrm{L}$ of solution. The pH of the solution is}

• A. 2
• B. 1
• C. 3
• D. 4

Hint:

pH = $-\log[H^+]$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
pH = $-\log[H^+]$.
Step 2: Detailed Explanation:
Moles of HCl = $\frac{0.192}{36.5} = 0.00526$ mol. Molarity = $\frac{0.00526}{0.5} = 0.01052$ M. $[H^+] = 0.01052 \times 0.95 = 0.01$ M. pH = $-\log(0.01) = 2$.
Step 3: Final Answer:
pH = 2.