Question

Calculate poH of 0.05M Ca(OH)2

Answer 1

Hello!

You need to dissociate

Ca (OH) ₂ -------------> Ca²⁺ + 2 OH⁻

2 x [OH] = 2 x 0.05 = 0.1 = $1 * 10^{-1}\: mol / L$

you calculate the pOH

$pOH = -\:log [OH]$

$pOH = -\;log [1 * 10^{-1}]$

$pOH = 1 - log\:1$

$pOH = 1 - 0$

$\boxed{\boxed{pOH = 1}}\end{array}}\qquad\checkmark$

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I Hope this helps, greetings ... DexteR! =)