(D) Given that $NaOH$ is a strong base,it dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$.Therefore,$[OH^-] = [NaOH] = 5.0 \times 10^{-2} \

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(D) Given that $NaOH$ is a strong base,it dissociates completely as: $NaOH \rightarrow Na^+ + OH^-$.Therefore,$[OH^-] = [NaOH] = 5.0 \times 10^{-2} \ M$.Now,calculate $pOH$:$pOH = -\log [OH^-] = -\log (5.0 \times 10^{-2})$$pOH = -(\log 5 + \log 10^{-2}) = -(\log (10/2) - 2)$$pOH = -(1 - \log 2 - 2) = -(-1 - 0.3) = 1.30$.Using the relation $pH + pOH = 14$ at $25^{\circ}C$:$pH = 14 - pOH = 14 - 1.30 = 12.70$.

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

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$NaOH$ is a strong base. What will be the $pH$ of a $5.0 \times 10^{-2} \ M$ $NaOH$ solution? (Given: $\log \ 2 = 0.3$)

JEE MAIN 2013 All JEE MAIN

A
$14$
B
$13.70$
C
$13$
D
$12.70$

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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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$NaOH$ is a strong base. What will be the $pH$ of a $5.0 \times 10^{-2} \ M$ $NaOH$ solution? (Given: $\log \ 2 = 0.3$)

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Calculate the $pH$ of $1.0 \times 10^{-8} \ M \ NaOH$ solution.

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