Select the correct option that has the same p | vedclass

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

Question

(C) To find the $pH$ of each solution:$(1) 0.005 \, M \, H_2SO_4$: Since $H_2SO_4$ is a strong diprotic acid,$[H^+] = 2 	imes 0.005 = 0.01 \, M = 10^

Vedclass · Accepted Answer

$1, 4$

Vedclass · Answer

(C) To find the $pH$ of each solution:$(1) 0.005 \, M \, H_2SO_4$: Since $H_2SO_4$ is a strong diprotic acid,$[H^+] = 2 	imes 0.005 = 0.01 \, M = 10^{-2} \, M$. Thus,$pH = -\log(10^{-2}) = 2$.$(2) 0.1 \, M \, Na_2SO_4$: This is a salt of a strong acid and a strong base,so it is neutral. $pH = 7$.$(3) 10^{-2} \, M \, NaOH$: Since $NaOH$ is a strong base,$[OH^-] = 10^{-2} \, M$. Thus,$pOH = -\log(10^{-2}) = 2$,and $pH = 14 - 2 = 12$.$(4) 0.01 \, M \, HCl$: Since $HCl$ is a strong monoprotic acid,$[H^+] = 0.01 \, M = 10^{-2} \, M$. Thus,$pH = -\log(10^{-2}) = 2$.Comparing the $pH$ values,solutions $(1)$ and $(4)$ have the same $pH$ of $2$.

Select the correct option that has the same $pH$ value.
$(1) 0.005 \, M \, H_2SO_4$
$(2) 0.1 \, M \, Na_2SO_4$
$(3) 10^{-2} \, M \, NaOH$
$(4) 0.01 \, M \, HCl$

Similar Questions

The ratio of $pH$ of $0.1 \, M \, H_2SO_4$ and that of $0.1 \, N \, H_2SO_4$ is

What is the $pH$ of a $10^{-4} \ M \ OH^{-}$ solution at $330 \ K$,if $K_w$ at $330 \ K$ is $10^{-13.6}$?

The $pH$ of a $0.02 \, M$ solution of hydrochloric acid is

How many moles of $HCl$ must be removed from $1 \ L$ of an aqueous $HCl$ solution to change its $pH$ from $2$ to $3$?

Water is added to $1.0 \ mL$ of $0.1 \ M$ $HCl$ solution to make the total volume $50 \ mL$. Calculate the $pH$ change of the solution.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Select the correct option that has the same $pH$ value.$(1) 0.005 \, M \, H_2SO_4$$(2) 0.1 \, M \, Na_2SO_4$$(3) 10^{-2} \, M \, NaOH$$(4) 0.01 \, M \, HCl$