Question #8a714

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Question #8a714

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Answer 1 · 2016-12-22T07:39:41

$p H = 12$ $pH=12$

Explanation:

$[H O^{-}] = \frac{4 \cdot g}{39.99 \cdot g \cdot m o l^{- 1}} \times \frac{1}{10 \cdot L} = 0.01 \cdot m o l \cdot L^{- 1}$ $[HO^-]=(4*g)/(39.99*g*mol^-1)xx1/(10*L)=0.01*mol*L^-1$ .

Now $p O H = - {log}_{10} [H O^{-}] = - {log}_{10} (0.01) = 2.0$ $pOH=-log_10[HO^-]=-log_10(0.01)=2.0$

But $p H + p O H = 14$ $pH+pOH=14$ , and thus $p H = 12$ $pH=12$

That $p H + p O H = 14$ $pH+pOH=14$ is key, and of course it stems from the autoprotolysis equilibrium, $[H O^{-}] [H_{3} O^{+}] = 10^{- 14}$ $[HO^-][H_3O^+]=10^-14$ .

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Question #8a714

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