1. Calculate values of K SP for the following salts using the information given:

ID: 825978 • Letter: 1

Question

1. Calculate values of KSP for the following salts using the information given:

a. A sample of 4.8 x 10-5 mol of calcium oxalate (CaC2O4) dissolves in 1.0 L of water to produce a saturated solution.

CaC2O4 (s) ? Ca2+ (aq) + C2O42- (aq)

b. The concentration of Pb2+ in a solution saturated with PbBr2 is 2.14 x 10-2 M.

c. The molar solubility of BiI3 is 1.32 x 10-5 mol / L.

2. The KSP for lead (II) iodide is 1.4 x 10-8. Calculate the solubility of lead (II) iodide in each of the following:

a. Water

b. 0.10 M lead (II) nitrate

c. 0.010 M sodium iodide

1.a.[Ca2+]=4.8*10^(-5)

Ksp=[Ca2+]^2=2.3*10^(-9)

b.[Pb2+]=2.14*10^(-2)

and [Br-]=2*[Pb2+]

Ksp=4[Pb2+]^3=4[2.14*10^(-2)]^3=3.9*10^(-5)

c.Ksp=[Bi3+][3I-]^3=[1.32*10^(-5)][3*1.32*10^(-5)]^3=2.7*10^(-19)

2.a.[Pb2+][2I-]^2=1.4*10^(-8)

4*[x]^3=1.4*10^(-8) (where x is the solubility in mol/L)

x=1.5*10^(-3) M

b.let solubility be x then

[Pb2+]=0.1+x and [I-]=2x

(0.1+x)(2x)^2=1.4*10^(-8)

neglecting x wrt O.1 since x<<0.1

4x^2=1.4*10^(-8)

x=1.36*10^(-4) M

c.similarly

[Pb2+][2I-]=1.4*10^(-8)

[I-]=0.01+2x

and [Pb2+]=x

then

x(0.01+2x)^2=1.4*10^(-8)

2x<<0.01

so 0.0001x=1.4*10^(-8)

x= 1.4*10^(-4) M

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