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What are the hydronium and hydroxide ion concentrations in a solution that has a pH of 12.330?

1 Answer

May 19, 2017

$[H^{+}] = 4.68 \times 10^{- 13} M$ $[H^+] = 4.68 xx 10^-13 M$
$[O H^{-}] = 0.0214 M$ $[OH^-] = 0.0214 M$

Explanation:

The $[H^{+}]$ $[H^+]$ can be found by raising $10$ $10$ to the power of the negative of the $p H$ $pH$ :

$[H^{+}] = 10^{- 12.330} = 4.68 \times 10^{- 13} M$ $[H^+] = 10^(-12.330) = 4.68 xx 10^-13 M$

The $[O H^{-}]$ $[OH^-]$ can be found by dividing the value of the ion-product constant for water, $K_{w}$ $K_w$ , by the $[H^{+}]$ $[H^+]$ :

$[O H^{-}] = \frac{1.00 \times 10^{- 14} M}{4.68 \times 10^{- 13} M} = 0.0214 M$ $[OH^-] = (1.00 xx 10^-14 M)/(4.68 xx 10^-13 M) = 0.0214 M$

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