Calculate the pH of a 0.018 , M sodium acetat | vedclass

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

Question

(A) Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.For such salts,the $pH$ is given by the formula: $pH

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.For such salts,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$.Given: $C = 0.018 \, M = 1.8 	imes 10^{-2} \, M$ and $K_a = 1.8 	imes 10^{-5}$.First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 	imes 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.255 = 4.745 \approx 4.75$.Now,substitute the values into the formula:$pH = 7 + \frac{1}{2}(4.75) + \frac{1}{2}\log(1.8 	imes 10^{-2})$$pH = 7 + 2.375 + \frac{1}{2}(\log 1.8 + \log 10^{-2})$$pH = 7 + 2.375 + \frac{1}{2}(0.255 - 2)$$pH = 7 + 2.375 + \frac{1}{2}(-1.745)$$pH = 7 + 2.375 - 0.8725 = 8.5025 \approx 8.5$.

Calculate the $pH$ of a $0.018 \, M$ sodium acetate solution,given its $K_a = 1.8 \times 10^{-5}$. (in $.5$)

Similar Questions

Do you expect different products in solution when aluminium $(III)$ chloride and potassium chloride are treated separately with $(i)$ normal water,$(ii)$ acidified water,and $(iii)$ alkaline water? Write equations wherever necessary.

The $pH$ of the solution obtained on neutralization of $40 \ mL \ 0.1 \ M \ NaOH$ with $40 \ mL \ 0.1 \ M \ CH_3COOH$ is

Which among the following salts undergoes hydrolysis?

What is the $pH$ of an aqueous solution of sodium acetate?

The $pK_a$ of a weak acid $(HA)$ and $pK_b$ of a weak base $(BOH)$ are $3.2$ and $3.4,$ respectively. The $pH$ of their salt $(AB)$ solution is

Vedclass Test Series

Exam Paper Generator

Online Exam Module