What is the pH of a $1.0 \cdot 10^{- 3}$ $1.0 * 10^-3$ $M$ $M$ sodium hydroxide solution?

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Answer 1 · 2016-12-28T05:55:13

$p H = 11$ $pH=11$

We examine the equilibrium:

$2 H_{2} O (l) ⇌ H_{3} O^{+} +^{-} O H$ $2H_2O(l) rightleftharpoons H_3O^+ + ""^(-)OH$

Under standard conditions, $[H_{3} O^{+}] [H O^{-}] = 10^{- 14}$ $[H_3O^+][HO^-]=10^-14$ .

And if we take ${log}_{10}$ $log_10$ of both sides:

${log}_{10} [H_{3} O^{+}] + {log}_{10} [H O^{-}] = - 14$ $log_10[H_3O^+]+log_10[HO^-]=-14$

Om rearrangement:

$14 = - {log}_{10} [H_{3} O^{+}] - {log}_{10} [H O^{-}]$ $14=-log_10[H_3O^+]-log_10[HO^-]$

But by definition, $- {log}_{10} [H_{3} O^{+}] = p H$ $-log_10[H_3O^+]=pH$ , and ${log}_{10} [H O^{-}] = p O H$ $log_10[HO^-]=pOH$ .

And thus our definining relationship: $p H + p O H = 14$ $pH+pOH=14$ .

Since (finally!) $p O H = - {log}_{10} [H O^{-}] = - {log}_{10} (1 \times 10^{- 3}) = - (- 3) = 3$ $pOH=-log_10[HO^-]=-log_10(1xx10^-3)=-(-3)=3$ .

And if $p O H = 3$ $pOH=3$ , $p H = 14 - 3 = ? ?$ $pH=14-3=??$

In A level, you do have to remember the result:

$p H + p O H = 14$ $pH+pOH=14$ . With this is mind, these problems become trivial.

What is the pH of a 1.0⋅10−31.0 * 10^-3 MM sodium hydroxide solution?