The pH of a 10^{-8} M solution of HCl in wat | vedclass

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

Question

(C) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected. The total concentration of $

Vedclass · Accepted Answer

between $6$ and $7$

Vedclass · Answer

(C) For a very dilute acid solution,the contribution of $H^+$ ions from the auto-ionization of water cannot be neglected. The total concentration of $H^+$ ions is given by: $[H^+]_{total} = [H^+]_{HCl} + [H^+]_{water} = 10^{-8} \ M + 10^{-7} \ M$. $[H^+]_{total} = (0.1 	imes 10^{-7} + 1 	imes 10^{-7}) \ M = 1.1 	imes 10^{-7} \ M$. Now,calculate the $pH$: $pH = -\log[H^+]_{total} = -\log(1.1 	imes 10^{-7})$. $pH = 7 - \log(1.1) \approx 7 - 0.0414 = 6.9586$. Thus,the $pH$ lies between $6$ and $7$.

The $pH$ of a $10^{-8} \ M$ solution of $HCl$ in water is

Similar Questions

$50 \ mL$ of $H_2O$ is added to $50 \ mL$ of $1 \times 10^{-3} \ M$ barium hydroxide solution. What is the $pH$ of the resulting solution?

The concentration in $M$ of $OH^{-}$ in $0.001 \ M$ $H_2SO_4$ is

The $pH$ of the solution obtained by mixing $100 \ mL$ of a solution of $pH=3$ with $400 \ mL$ of a solution of $pH=4$ is

Equal volumes of two solutions of $pH = 3$ and $pH = 5$ are mixed. The resultant $pH$ is

What is the $pH$ of $0.005 \ M \ H_2SO_4$ solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module