Question

The Ks of strontium sulfite, SrSO3, is 4.0 x 10^−8 at 25°C. If 1.0000 g of SrSO3 is stirred in 5.0 L of water until the solution is saturated and then filtered, What mass of SrSO3 would remain?

Answer 1

Molar mass = 183.68 gms

Concentration of salt=$c=1/(183.68*5)= 1.088*10^{-3}M$

$K_{sp} = s^2=4*10^{-8}M \implies s= *10^{-4}M = 0.2*10^{-3}M$

Thus, concentration of SrSO3 remaining is $c-s= 0.888*10^{-3}M$

Moles remaining = $0.888*10^{-3}*5=4.444*10^{-3} mol$

Thus, amount of SrSO3 remaining = $4.444*10^{-3}*183.68=0.8163 gms.$