If the pKa of lactic acid is 5,then the pH of | vedclass

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

Question

(C) Calcium lactate is the salt of a weak acid (lactic acid) and a strong base $(Ca(OH)_2)$.Formula for $pH$ of a salt of weak acid and strong base is

Vedclass · Accepted Answer

$83$

Vedclass · Answer

(C) Calcium lactate is the salt of a weak acid (lactic acid) and a strong base $(Ca(OH)_2)$.Formula for $pH$ of a salt of weak acid and strong base is:$pH = 7 + \frac{1}{2}(pKa + \log C)$Here,$C$ is the concentration of the salt,which is $0.005 \ M$.$pH = 7 + \frac{1}{2}(5 + \log(0.005))$$pH = 7 + \frac{1}{2}(5 + \log(5 	imes 10^{-3}))$$pH = 7 + \frac{1}{2}(5 + \log 5 - 3)$$pH = 7 + \frac{1}{2}(2 + 0.699) = 7 + 1.3495 = 8.3495$Rounding to the nearest integer for the form $x 	imes 10^{-1}$,we get $83.495 \approx 83 	imes 10^{-1}$.

If the $pKa$ of lactic acid is $5$,then the $pH$ of $0.005 \ M$ calcium lactate solution at $25^{\circ} \ C$ is $........ \times 10^{-1}$ (Nearest integer). The structure of lactic acid is given below:

Similar Questions

The degree of hydrolysis of a salt of weak acid and weak base in its $0.1 \, M$ solution is found to be $50 \%$. If the molarity of the solution is $0.2 \, M$,the percentage hydrolysis of the salt should be $...... \%$.

The $pH$ of a $0.02 \ M \ NH_4Cl$ solution will be [given $K_b \ (NH_4OH) = 10^{-5}$ and $\log \ 2 = 0.301$ ]

$A$ $100 \ mL$, $0.1 \ M$ solution of ammonium acetate is diluted by adding $100 \ mL$ of water. The pH of the resulting solution will be ($pK_{a}$ of acetic acid is nearly equal to $pK_{b}$ of $NH_{4}OH$)

The aqueous solution of $CuSO_4$ is

Which salt from the following forms an aqueous solution having $pH$ less than $7$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module