Question #4af57

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Question #4af57

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Answer 1 · 2015-01-12T08:22:24

Sodium hydroxide is a strong base, which means it dissociates completely in aqueous solution into ${Na}^{+}$ $"Na"^(+)$ and ${OH}^{-}$ $"OH"^(-)$ ions.

This tells you that the concentrations of the ${Na}^{+}$ $"Na"^(+)$ cations and of the ${OH}^{-}$ $"OH"^(-)$ anions will be equal to each other, and to the initial concentration of $NaOH$ $"NaOH"$ .

Since you start with $10^{- 5}$ $10^(-5)$ $M$ $"M"$ $NaOH$ $"NaOH"$ , in aqueous solution you'll end up with

$[{Na}^{+}] = 10^{- 5}$ $["Na"^(+)] = 10^(-5)$ $M$ $"M"$ and $[{OH}^{-}] = 10^{- 5}$ $["OH"^(-)] = 10^(-5)$ $M$ $"M"$

The equation to use in this case is

$p O H = - log ([O H^{-}]) = - log (10^{- 5}) = 5$ $pOH = -log([OH^(-)]) = -log(10^(-5)) = 5$

This implies a pH of

$p H = 14 - p O H = 14 - 5 = 9$ $pH = 14 - pOH = 14 - 5 =9$

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Question #4af57

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