Question

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

Answer 1

`s= 5.9*10^(-4) M`

Step-by-step 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`