Assuming that no equilibria other than dissolution are involved, calculate the m

Question

Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) Ag2SO4 (b) PbBr2 (c) AgI (d) CaC2O4H2O Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) Ag2SO4 (b) PbBr2 (c) AgI (d) CaC2O4H2O Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) Ag2SO4 (b) PbBr2 (c) AgI (d) CaC2O4H2O

Explanation / Answer

1.Molar solubility of Ag2SO4 -

Ag2SO4 ---> 2Ag+ + SO42-

... x ............ 2x ......... x ......

Ksp of Ag2SO4 is1.2*10^-5

Here x is the molar solubility

Ksp = [Ag+]^2*[SO4 2-] = (2x)^2*x = 4x^3

1.2*10^-5 = 4x^3

x = 1.44 x10^-2M

Molar solubility = 1.44 x 10^-2 M

Ag2SO4 [molar mass = 312g/mol]

Molar solubility of Ag2SO4 in g/L =1.44 x 10^-2 M x 312g/mol

                                                          = 4.49g/L

2.Molar solubility of PbBr2

The Ksp of PbBr2 is 6.60× 10^–6

PbBr2---> Pb2+ + 2Br –

x            x          2x

Ksp = [Pb2+] [Br-]^2

6.60× 10^–6=[x][2x]^2

6.60× 10^–6 =4x^3

So, x = 1.16 x 10^-2M

Molar solubility = 1.16 x 10^-2 M

Molar mass of PbBr2 = 367g/mol

Molar solubility of PbBr2 in g/L =1.16 x 10^-2 M x 367g/mol

                                                          = 4.26g/L

3.Molar solubility of AgI

Ksp of AgI is 1.5 x 10-16.

AgI ---------> Ag+ + I-

x                       x       x

Ksp = [Ag+][I-]

1.5 x 10^-16 = [x] [x]

1.5 x 10^-16 = x^2

x = 1.22 x 10^-8

Molar solubility = 1.22 x 10^-8 M

Molar mass of AgI = 234.77g/mol

Molar solubility of AgI in g/L = 1.22 x 10^-8 M x 234.77g/mol = 2.86 x 10^-6 g/L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---