(C) $NaOH \longrightarrow Na^{+} + OH^{-}$ Since $NaOH$ is a strong base,$[OH^{-}] = [NaOH] = 0.005 \ M = 5 \times 10^{-3} \ M$. $pOH = -\log_{10}[OH^

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(C) $NaOH \longrightarrow Na^{+} + OH^{-}$ Since $NaOH$ is a strong base,$[OH^{-}] = [NaOH] = 0.005 \ M = 5 \times 10^{-3} \ M$. $pOH = -\log_{10}[OH^{-}]$ $pOH = -\log_{10}(5 \times 10^{-3}) = -(\log_{10} 5 + \log_{10} 10^{-3})$ $pOH = -(0.699 - 3) = -(-2.301) = 2.301$. We know that $pH + pOH = 14$ at $298 \ K$. $pH = 14 - 2.301 = 11.699 \approx 11.7$.

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 $0.005 \ M \ NaOH$ solution?

MHT CET 2023 All MHT CET

A
$2.30$
B
$12.6$
C
$11.7$
D
$3.2$

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What is the $pH$ of $0.005 \ M \ NaOH$ solution?

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