What is the pH of a $6.49 \cdot 10^{- 3}$ $6.49 * 10^-3$ $M$ $M$ $K O H$ $KOH$ solution?

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What is the pH of a $6.49 \cdot 10^{- 3}$ $6.49 * 10^-3$ $M$ $M$ $K O H$ $KOH$ solution?

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Answer 1 · 2016-05-28T14:11:03

$p H ≅ 11.8$ $pH~=11.8$

Explanation:

It is a fact that in aqueous solution under standard conditions,

$14 = p H + p O H$ $14=pH+pOH$

So we can directly calculate $p O H$ $pOH$ , and then approach $p H$ $pH$ .

$p O H$ $pOH$ $=$ $=$ $- {log}_{10} [H O^{-}]$ $-log_10[HO^-]$ $=$ $=$ $- {log}_{10} [6.49 \times 10^{- 3}]$ $-log_10[6.49xx10^-3]$ $=$ $=$ $2.19$ $2.19$

Thus $p H$ $pH$ $=$ $=$ $14.00 - 2.19$ $14.00-2.19$ $=$ $=$ $? ?$ $??$

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