Question

Determine the mass of grams of led(II) sulfate that will dissolve in 2.50 x 10^2 ml water. Ksp= 1.8 x 10^-8

MM=303.3 g/mol

Answer 1

9.86*10^(-3) g

Step-by-step explanation:

PbSO4 ----> Pb^(2+) + SO4^(2-)

s s

Ksp = s²

s =√Ksp = √(1.8*10^-8) = 1.3*10^(-4) mol/L

The molar solubility PbSO4 = 1.3*10^(-4) mol/L.

2.50 *10^2 mL *1L/10³mL =0.250L

1.3*10^(-4)mol/L *0.250L*303.3 g/mol = 9.86*10^(-3) g