Write the equilibrium expression the dissolution of solid lead(Il) iodide, Pbl2

Question

Write the equilibrium expression the dissolution of solid lead(Il) iodide, Pbl2 into water. Write the equation relating the Solubility Product, K_sp for Pbl_2 to the molar solubility, S, of the compound. 50.00 mL of a standard solution of lead(II) iodide in water was passed through an ion-exchange resin (in the acid form). the protons released were collected and titrated against a standard solution of 0.008180 M NaOH. the endpoint of the titration was 18 57 mL. Calculate: the Molar Solubility, S of Pbl_2; and the value of the Solubility Product, K_sp for lead(II) iodide.

Explanation / Answer

(a) PbI2 <---> Pb2+ + 2I-

(b) If molar solubility is S, then:

Ksp = [Pb2+][2I-]2 = S.(2S)2 = 4S3

(c) In the ion exchange resin, one mole of Pb2+ will replace 2 moles of H+.

The released protons will be titrated by NaOH.

No of moles of NaOH needed to titrate the protons

= 0.008180*18.57/1000

= 1.519*10-4 moles

As 1 mole of NaOH titrates 1 mole of H+ so amount of H+ released = 1.519*10-4 moles

Amount of Pb2+ in 50 mL solution = 1.519*10-4 moles/2 = 7.595*10-5 moles

Concentration of lead ions = (7.595*10-5 ) * 1000/50 = 1.519*10-3 M

(i)            PbI2        <--->      Pb2+ + 2I-

               S                            S             2S          

The standard solution of PbI2 gives 1.519*10-3 M lead ions in solution. So S = 1.519*10-3 M

Molar solubility of PbI2, S = 1.519*10-3 M

(ii) Ksp = [Pb2+][2I-]2 = S.(2S)2 = 4S3

= 4 (1.519*10-3)3

= 1.402*10-8

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