Calculate $p H$ $pH$ of $10^{- 9}$ $10^-9$ $M$ $M$ $N a O H$ $NaOH$ ?

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Answer 1 · 2016-01-30T17:26:54

$p H = 7$ $pH=7$

Consider the two equilibria happening in solution:

$N a O H ⇌ N a^{+} + O H^{-}$ $NaOHrightleftharpoonsNa^(+)+OH^-$

and

$H_{2} O ⇌ H^{+} + O H^{-}$ $H_2OrightleftharpoonsH^(+)+OH^-$

From $N a O H$ $NaOH$ : $[O H^{-}] = 10^{- 9} M$ $[OH^-]=10^(-9)M$

From $H_{2} O$ $H_2O$ : $[O H^{-}] = 10^{- 7} M$ $[OH^-]=10^(-7)M$

Since ${[O H^{-}]}_{H_{2} O}^{-} >> {[O H^{-}]}_{N a O H}^{-}$ $[OH^-]_(H_2O)" >> "[OH^-]_(NaOH)$ ,

The $p O H$ $pOH$ will be determined from the ${[O H^{-}]}_{H_{2} O}^{-} = 10^{- 7} M$ $[OH^-]_(H_2O)=10^(-7)M$

$p O H = - log ([O H^{-}]) = - log (10^{- 7}) = 7$ $pOH=-log([OH^-])=-log(10^(-7))=7$

Therefore, to calculate the $p H$ $pH$ :

$p H = p K_{w} - p O H = 14 - 7 = 7$ $pH=pK_w-pOH=14-7=7$

Here is a video that explains in details this topic and treats a similar example:

Acids & Bases | pH of a Strong Acid & a Strong Base.

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