Calculate pH and pOH of 0.03 M NaOH solution | vedclass

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(B) $NaOH$ is a strong base and dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$. Concentration of $[OH^-] = 0.03 \ M$. $pOH = -\log[OH^-] =

Vedclass · Accepted Answer

$12.48$ and $1.52$

Vedclass · Answer

(B) $NaOH$ is a strong base and dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$. Concentration of $[OH^-] = 0.03 \ M$. $pOH = -\log[OH^-] = -\log(0.03) = -(\log 3 + \log 10^{-2}) = -(0.477 - 2) = 1.523 \approx 1.52$. Since $pH + pOH = 14$ at $298 \ K$,$pH = 14 - 1.52 = 12.48$. Therefore,the $pH$ is $12.48$ and the $pOH$ is $1.52$.

Calculate $pH$ and $pOH$ of $0.03 \ M$ $NaOH$ solution.

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