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Question #7c9bd

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Answer 1 · 2014-12-13T10:08:35

The answer is pH = $11.97$ $11.97$ .

The balanced chemical equation is

$K O H_{(s)} + H_{2} O_{(l)} \to K_{(a q)}^{+} + O H_{(a q)}^{-}$ $KOH_((s)) + H_2O_((l)) -> K_((aq))^+ + OH_((aq))^-$

Since $K O H$ $KOH$ is a strong base, it will completely dissociate into $K^{+}$ $K^+$ and $O H^{-}$ $OH^-$ ions; this means that the concentration of both $K^{+}$ $K^+$ and $O H^{_}$ $OH^_$ will be equal to the starting concentration of $9.28 \cdot 10^{- 3} M$ $9.28 * 10^-3M$ .

We know that pH = $14 - p O H$ $14 - pOH$ , which means that we need to determine the **pOH ** first

$p O H = - log ([O H^{-}]) = - log (9.28 \cdot 10^{- 3}) = 2.03$ $pOH = -log([OH^-]) = -log(9.28 * 10^-3) = 2.03$

Therefore, the pH is equal to $p H = 14 - 2.03 = 11.97$ $pH = 14 - 2.03 = 11.97$

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