NaOH is a strong base. What will be the pH of | vedclass

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

Vedclass · Accepted Answer

$12.70$

Vedclass · Answer

(D) Given that $NaOH$ is a strong base,it dissociates completely as: $NaOH ightarrow Na^+ + OH^-$.Therefore,$[OH^-] = [NaOH] = 5.0 	imes 10^{-2} \ M$.Now,calculate $pOH$:$pOH = -\log [OH^-] = -\log (5.0 	imes 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 $25^{\circ}C$:$pH = 14 - pOH = 14 - 1.30 = 12.70$.

$NaOH$ 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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