Number of equivalents of H_2SO_4 present in 1 | vedclass

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

Question

(B) Given $pH = 5$. Since $pH = -\log[H^+]$,we have $[H^+] = 10^{-5} \ M$. For $H_2SO_4$,which is a strong diprotic acid,the normality $N$ is equal to

Vedclass · Accepted Answer

$10^{-6}$

Vedclass · Answer

(B) Given $pH = 5$. Since $pH = -\log[H^+]$,we have $[H^+] = 10^{-5} \ M$. For $H_2SO_4$,which is a strong diprotic acid,the normality $N$ is equal to the molarity $M$ multiplied by the $n$-factor $(n=2)$. However,$[H^+] = N$ for a strong acid solution. Thus,$N = [H^+] = 10^{-5} \ N$. Volume of solution $V = 100 \ mL = 0.1 \ L$. Number of equivalents = $N 	imes V(L) = 10^{-5} 	imes 0.1 = 10^{-6} \ 	ext{equivalents}$.

Number of equivalents of $H_2SO_4$ present in $100 \ mL$ of its solution whose $pH$ is $5$.

Similar Questions

The $pH$ of $1 \ L$ of an aqueous solution of $HCl$ is $1$. How many liters of water must be added to make the $pH$ of this solution $2$ (in $L$)?

What is the pH of $10^{-8} \ M \ HCl$ solution?

$1 \, mL$ of $0.1 \, M$ $HCl$ is added to $999 \, mL$ of $0.1 \, N$ $NaCl$ solution. The $pH$ of the resulting solution will be:

$0.023 \ g$ of sodium metal is reacted with $100 \ cm^{3}$ of water. The $pH$ of the resulting solution is

Calculate the $pH$ of a $0.002 \ M$ $KOH$ solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module