Question

The pH of 0.1 M solution of monobasic acid is 2.34. Calculate the degree of dissociation of the acid.

• A. $3.1 \times 10^{-2}$
• B. $4.5 \times 10^{-2}$
• C. $2.18 \times 10^{-2}$
• D. $2.5 \times 10^{-3}$

Hint:

If the pH were exactly 2, $[\text{H}^+]$ would be $10^{-2} = 0.01$, making $\alpha = \frac{0.01}{0.1} = 0.1$. Because the pH is slightly higher at 2.34, the acidity is lower, meaning $\alpha$ must be somewhat smaller than $0.1$. The value $0.045$ ($4.5 \times 10^{-2}$) fits perfectly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the molar concentration ($C$) and the measured pH of a weak monobasic acid solution. We need to evaluate its degree of dissociation ($\alpha$).

Step 2: Key Formula or Approach:
For a weak monobasic acid dissociating in water ($\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$), the concentration of hydrogen ions relates to the degree of dissociation via: $$ [\text{H}^+] = \alpha \times C \implies \alpha = \frac{[\text{H}^+]}{C} $$ The hydrogen ion concentration can be recovered from the pH using the definition: $$ \text{pH} = -\log_{10}[\text{H}^+] \implies [\text{H}^+] = 10^{-\text{pH}} = \text{antilog}(-\text{pH}) $$

Step 3: Detailed Explanation:
Let's first determine the concentration of $\text{H}^+$ ions from the given $\text{pH} = 2.34$: $$ \log_{10}[\text{H}^+] = -2.34 $$ To convert this to standard characteristics and mantissa form for antilog evaluation: $$ -2.34 = -2 - 0.34 = (-2 - 1) + (1 - 0.34) = -3 + 0.66 = \bar{3}.66 $$ $$ [\text{H}^+] = \text{antilog}(\bar{3}.66) \approx 4.57 \times 10^{-3}\ \text{mol dm}^{-3} $$ Now, substitute the value of $[\text{H}^+]$ and the analytical concentration $C = 0.1\ \text{M}$ into the dissociation equation: $$ \alpha = \frac{4.57 \times 10^{-3}}{0.1} = 4.57 \times 10^{-2} $$ Rounding to match the closest given multiple-choice values yields $4.5 \times 10^{-2}$.

Step 4: Final Answer: The degree of dissociation of the weak monobasic acid is $4.5 \times 10^{-2}$, matching option (B).