What is the pOH of a solution with a pH of .7?

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Answer 1 · 2016-12-11T07:48:01

And it's an aqueous solution?

We know that in water the following equilibrium operates:

$2 H_{2} O (l) ⇌ H_{3} O^{+} + H O^{-}$ $2H_2O(l) rightleftharpoonsH_3O^+ + HO^-$ $K_{eq} = 10^{- 14}$ $K_"eq"=10^-14$

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

We take ${log}_{10}$ $log_10$ of both sides:

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

${log}_{10} [H_{3} O^{+}] + {log}_{10} [H O^{-}] = - 14$ $log_10[H_3O^+]+log_10[HO^-]=-14$ , and multiply both sides by $- 1$ $-1$ ,

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

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

And thus $p H + p O H = 14$ $pH+pOH=14$ (for water under standard conditions). This is something that you have to know.

You have $p H = 0.7$ $pH=0.7$ , and thus $p O H = 13.3$ $pOH=13.3$ . What is $[H_{3} O^{+}]$ $[H_3O^+]$ ?