(D) For a solution,$pH = -\log[H^{+}]$.Given $pH = 1$,so $[H^{+}] = 10^{-1} = 0.1\,M$.For $H_2SO_4$,the dissociation is $H_2SO_4 \rightarrow 2H^{+} +

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(D) For a solution,$pH = -\log[H^{+}]$.Given $pH = 1$,so $[H^{+}] = 10^{-1} = 0.1\,M$.For $H_2SO_4$,the dissociation is $H_2SO_4 \rightarrow 2H^{+} + SO_4^{2-}$.Since $1\,mol$ of $H_2SO_4$ produces $2\,mol$ of $H^{+}$,the concentration of $H^{+}$ is $2 \times [H_2SO_4]$.Therefore,$0.1 = 2 \times [H_2SO_4]$.$[H_2SO_4] = \frac{0.1}{2} = 0.05\,M$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)pH of strong Acids and strong Bases

Easy

Assuming complete ionisation,the $pH$ of $0.1\,M\,HCl$ is $1$. The molarity of $H_2SO_4$ with the same $pH$ is (in $,M$)

A
$0.2$
B
$0.1$
C
$2$
D
$0.05$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

Calculate the volume of $0.1 \ M \ NaOH$ required to complete the neutralization of $300 \ mL$ of $HCl$ solution having $pH = 2.25$. (in $mL$)

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The $pH$ of $100 \ mL$ of an aqueous solution of a strong acid is $1$. The volume of water to be added to change its $pH$ to $2$ is .......... $mL$.

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What is the $pH$ of $1 \text{ M } HCl$ solution? (Assume complete dissociation)

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

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Assuming complete ionisation,the $pH$ of $0.1\,M\,HCl$ is $1$. The molarity of $H_2SO_4$ with the same $pH$ is (in $,M$)

Similar Questions

Calculate the volume of $0.1 \ M \ NaOH$ required to complete the neutralization of $300 \ mL$ of $HCl$ solution having $pH = 2.25$. (in $mL$)

If $4.0 \, g$ of $NaOH$ is present in $1 \, L$ of solution,then its $pH$ will be:

The $pH$ of $100 \ mL$ of an aqueous solution of a strong acid is $1$. The volume of water to be added to change its $pH$ to $2$ is .......... $mL$.

What is the $pH$ of $1 \text{ M } HCl$ solution? (Assume complete dissociation)

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

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