If the pH=12 for a solution of potassium hydroxide, what is [HO^-]?

1 Answer
anor277
Oct 2, 2016

[HO^-] = 0.01*mol*L^-1.

Explanation:

We know, or should know that (i) pH=-log_10[H_3O^+], and pOH=-log_10[HO^-]; and (ii), given that K_w=[H_3O^+][""^(-)OH]=10^-14 under standard conditions, pK_w=pH+pOH=14.

We have pH=-log_10[H_3O^+]=12; thus [H_3O^+]=10^-12*mol*L^-1, and pOH=2, and thus [HO^-]=10^-2*mol*L^-1.

What is [K^+(aq)]?

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