(D) $NaOH$ is a strong base,so it dissociates completely in water: $[NaOH] = [OH^-] = 0.02 \ M = 2 \times 10^{-2} \ M$. $pOH = -\log[OH^-] = -\log(2 \

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(D) $NaOH$ is a strong base,so it dissociates completely in water: $[NaOH] = [OH^-] = 0.02 \ M = 2 \times 10^{-2} \ M$. $pOH = -\log[OH^-] = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.301 = 1.699 \approx 1.7$. Using the relation $pH + pOH = 14$ at $25^{\circ}C$: $pH = 14 - pOH = 14 - 1.7 = 12.3$.

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

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What is the $pH$ of $0.02 \ M \ NaOH$ solution?

MHT CET 2021 All MHT CET

A
$10.3$
B
$11.3$
C
$11.7$
D
$12.3$

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6-2.Equilibrium-II (Ionic Equilibrium)

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pH of strong Acids and strong Bases

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What is the $pH$ of $0.02 \ M \ NaOH$ solution?

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