(A) Step $1$: Calculate $[H^+]$ for solution $A$. Since $pH_A = 4$, $[H^+]_A = 10^{-pH_A} = 10^{-4} \text{ M}$.Step $2$: Calculate $[H^+]$ for solutio

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(A) Step $1$: Calculate $[H^+]$ for solution $A$. Since $pH_A = 4$, $[H^+]_A = 10^{-pH_A} = 10^{-4} \text{ M}$.Step $2$: Calculate $[H^+]$ for solution $B$. Given $[H^+]_B = 3 \times [H^+]_A = 3 \times 10^{-4} \text{ M}$.Step $3$: Calculate $pH_B$. $pH_B = -\log[H^+]_B = -\log(3 \times 10^{-4}) = -(\log 3 + \log 10^{-4}) = -(0.4771 - 4) = 4 - 0.4771 = 3.5229$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Water hydrolysis and pH scale

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There are two different solutions, $A$ and $B$. The $pH$ of solution $A$ is $4$. Find the $pH$ of solution $B$ having $[H^+]$ concentration three times that of solution $A$.

MHT CET 2026 All MHT CET

A
$3.5229$
B
$4.4771$
C
$3.4771$
D
$4.5229$

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6-2.Equilibrium-II (Ionic Equilibrium)

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There are two different solutions, $A$ and $B$. The $pH$ of solution $A$ is $4$. Find the $pH$ of solution $B$ having $[H^+]$ concentration three times that of solution $A$.

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Statement-$I$ :- $pH$ of pure water increases with increase in temperature.
Statement-$II$ :- Degree of ionisation increases with increase in temperature.