Answer to Question #107828 in General Chemistry for camila

Question

Calculate the pH of 1.0 x 10^-10 M Ca(OH)2

Answer 1

$[OH^-]=2[Ca^{2+}]+[H^+]$

$K_w=[H +] [OH^-]=([OH^-]-2[Ca^{2+}])[OH^-]$

$[OH^-]^2-2[Ca^{2+}][OH^-]-K_w=0$

$[OH^-]^2-2*10^{-10}[OH^-]-10^{-14}=0$

$[OH]=1.001*10^{-7}$

$pOH=-lg[OH^-]=6.999$

$pH=K_w-pOH=14-6.999=7.001$

The answer is 7.001