Question

Compute the moles/liter solubility of calcium hydroxide in 0.047 M sodium hydroxide. Ksp= 1.3 x 10^-6

Answer 1

Answer:

$s= 5.9*10^{-4} M$

Explanation:

Solubility is s.

$Ca(OH)_{2} --> Ca^{2+} +2OH^{-}$

                           s           2s

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

$K_{sp} = s*(2s)^{2}$

When we have Ca(OH)2 in Na(OH) solution, because NaOH has OH- also :

$K_{sp}=s*(2s+0.047)^{2}$,

because 2s << 0.047,

we can write

$K_{sp} = s*0.047^{2}\\\\1.3*10^{-6}=s*0.047^{2}\\\\s= 5.9*10^{-4} M$