In which of the following mixed aqueous solut | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The $pH$ of an acidic buffer solution is given by the Henderson-Hasselbalch equation:$pH = pK_{a} + \log \frac{[salt]}{[acid]}$For $pH = pK_{a}$,

Vedclass · Accepted Answer

Both $(1)$ and $(2)$ are correct

Vedclass · Answer

(D) The $pH$ of an acidic buffer solution is given by the Henderson-Hasselbalch equation:$pH = pK_{a} + \log \frac{[salt]}{[acid]}$For $pH = pK_{a}$, we must have $\log \frac{[salt]}{[acid]} = 0$, which implies $[salt] = [acid]$.$(1)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $100 \ mL$ of $0.1 \ M \ CH_{3}COONa$ $(10 \ mmol)$.Here, $[salt] = [acid] = 10 \ mmol$. Thus, $pH = pK_{a} + \log(1) = pK_{a}$.$(2)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $50 \ mL$ of $0.1 \ M \ NaOH$ $(5 \ mmol)$.The reaction is: $CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$.Initial: $10 \ mmol \ CH_{3}COOH, 5 \ mmol \ NaOH$.Final: $5 \ mmol \ CH_{3}COOH, 5 \ mmol \ CH_{3}COONa$.Since $[salt] = [acid] = 5 \ mmol$, $pH = pK_{a} + \log(1) = pK_{a}$.$(3)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $100 \ mL$ of $0.1 \ M \ NaOH$ $(10 \ mmol)$.This results in complete neutralization to form $CH_{3}COONa$. No buffer is formed.$(4)$ $CH_{3}COOH + NH_{3} \rightarrow CH_{3}COONH_{4}$.This forms a salt of a weak acid and a weak base, not a buffer solution.Therefore, both $(1)$ and $(2)$ satisfy the condition $pH = pK_{a}$.

Similar Questions

If $0.1 \, M, 100 \, mL$ $NaOH$ solution is mixed with $0.05 \, M, 100 \, mL$ $H_2SO_4$ solution,the $pH$ of the resulting solution is:

The $p^H$ of a mixture containing $10 \ mL$ of $0.1 \ N$ $NaOH$ and $10 \ mL$ of $0.05 \ N$ $H_2SO_4$ is ............

On decreasing the $pH$ from $7$ to $2$,the solubility of a sparingly soluble salt $(MX)$ of a weak acid $(HX)$ increased from $10^{-4} \ mol \ L^{-1}$ to $10^{-3} \ mol \ L^{-1}$. The $pK_{a}$ of $HX$ is:

When dilute $H_2SO_4$ is added to an aqueous solution of potassium chromate,the yellow color of the solution changes to orange. This indicates that .....

At $25 \, ^\circ C$,the dissociation constant of pure water is equal to .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module