The $K_a$ of $CH_3COOH$ is $1.8 \times 10^{-5}$. How many grams of $CH_3COONa$ are required in $0.1 \ M \ CH_3COOH$ to form a solution having $pH = 4.0$? (Molecular mass of $CH_3COONa = 82 \ g \ mol^{-1}$)
(A) Using the Henderson-Hasselbalch equation for an acidic buffer: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given $pH = 4.0$ and $K_a = 1.8 \times 10^{-5}$,we find $pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Substituting the values: $4.0 = 4.74 + \log \frac{[CH_3COONa]}{0.1}$.
$-0.74 = \log \frac{[CH_3COONa]}{0.1}$.
$\frac{[CH_3COONa]}{0.1} = 10^{-0.74} \approx 0.182$.
$[CH_3COONa] = 0.182 \times 0.1 = 0.0182 \ M$.
Mass of $CH_3COONa = \text{Molarity} \times \text{Volume} \times \text{Molar Mass}$. Assuming $1 \ L$ solution: $0.0182 \ mol \times 82 \ g \ mol^{-1} = 1.4924 \ g$.
Given $pH = 4.0$ and $K_a = 1.8 \times 10^{-5}$,we find $pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Substituting the values: $4.0 = 4.74 + \log \frac{[CH_3COONa]}{0.1}$.
$-0.74 = \log \frac{[CH_3COONa]}{0.1}$.
$\frac{[CH_3COONa]}{0.1} = 10^{-0.74} \approx 0.182$.
$[CH_3COONa] = 0.182 \times 0.1 = 0.0182 \ M$.
Mass of $CH_3COONa = \text{Molarity} \times \text{Volume} \times \text{Molar Mass}$. Assuming $1 \ L$ solution: $0.0182 \ mol \times 82 \ g \ mol^{-1} = 1.4924 \ g$.
Explore More
Similar Questions
$A$ certain buffer solution contains equal concentrations of $X^{-}$ and $HX$. The $K_a$ for $HX$ is $10^{-8}$. The $pH$ of the buffer is
Medium
View SolutionUnder which condition does a basic buffer follow the equation $pOH - pK_b = 1$?
Medium
View Solution$A$ solution is prepared by mixing $0.01 \ mol$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$ in $100 \ mL$ of water. The $pH$ of the resulting solution is. . . . . . .
[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]
[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]
MediumIIT 2022
View SolutionWhich buffer solution out of the following will have $pH > 7$?
Easy
View Solution$A$ student needs to prepare a buffer solution of propanoic acid and its sodium salt with $pH$ $4$. The ratio of $\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$ required to make the buffer is .....
Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$
Given: $K_{a}(CH_{3}CH_{2}COOH) = 1.3 \times 10^{-5}$
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo