Question

Determine the mass in grams of lead (II) iodide that will dissolve in 500.0 mL of a solution containing 1.82 grams of lead (II) nitrate. Ksp of lead (II) iodide is 1.4 x 10^-8.

Answer 1

Given,

Mass of lead(II)nitrate = 1.82 grams

K_sp of lead(II)iodide = 1.4 x 10^-8

Calculating the concentration of 1.82 g of lead(II)nitrate in 500.0 mL of a solution,

= 1.82 g of Pb(NO₃)₂ x (1 mol /331.2 g) x ( 1000mL / 1L) x (1 /500 mL)

= 0.01099 M of Pb(NO₃)₂

Now, The concentration of [Pb²⁺] = 0.01099 M

K_sp expression for lead(II)iodide,

PbI₂(s) Pb²⁺(aq) + 2I^-(aq)

	Pb2+(aq)	2I-(aq)
I(M)	0.01099	0
C(M)	+S	+2S
E(M)	0.01099+S	2S

Now, the K_sp expression is,

K_sp = [Pb²⁺] [I^-]²

1.4 x 10^-8 = [0.01099+S] [2S]²

S = 0.0005643 mol/L

Thus, the solubility of lead(II)iodide in a solution containing 1.82 grams of lead(II)nitrate is 0.0005643 mol/L

Now, the moles of lead(II)iodide dissolve in 500.0 mL of a solution,

= (0.0005643 mol / 1L) x (1L /1000 mL) x (500.0 mL)

= 0.0002822 mol of PbI₂

Converting the moles to grams,

= 0.0002822 mol of PbI₂ x (461.01 g /1 mol)

= 0.130 g of PbI₂