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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