(C) $NaOH_{(aq)}$ is a strong base solution. Since $NaOH$ dissociates completely,$[OH^-] = 0.01 \ M = 10^{-2} \ M$. $pOH = -\log[OH^-] = -\log(10^{-2}

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(C) $NaOH_{(aq)}$ is a strong base solution. Since $NaOH$ dissociates completely,$[OH^-] = 0.01 \ M = 10^{-2} \ M$. $pOH = -\log[OH^-] = -\log(10^{-2}) = 2$. We know that $pH + pOH = 14$ at $25^{\circ}C$. Therefore,$pH = 14 - pOH = 14 - 2 = 12$.

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

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The $pH$ of $0.01 \ M \ NaOH_{(aq)}$ solution will be

NEET 2019 All NEET

A
$7.01$
B
$2$
C
$12$
D
$9$

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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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The $pH$ of $0.01 \ M \ NaOH_{(aq)}$ solution will be

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