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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

Respuesta :

SvetkaChem

Answer:

9.86*10^(-3) g

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

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