Question

Explain, with detailed calculations, what will happen to the molar solubility of Pbl, (Ksp = 7.1 x 10-9) in a 0.35 M Bel2 (This is 100% soluble) solution.

Answer 1

Pb Iq is sparingly soluble salt in water. 2 Pb Iq (6) 2 Pb2+ (ar) + 2 I (a) a S S : 25 Let a's! be the molar solubility of pb. Is in water- The soluble Pb. In generaters Is' molar Pb2t and as' molar I ions in solution. (Molar solubility = mollut) . Expression for Ksp of pb Iz : i Ksp = [Pb2+] [I]? = 5.(25) = Äs? Again Ksp for. pb I2 = 7.1x100 Therefore, 4s3 = 71 x 10 9 or, s3 = 7.1 8109 S3 (7x10 mg * L ass=1:21. x 103 - Solubility of Pb Iz in water is 1.21 x 103 mollüz. Agains. Pb Iz is taken in a 0.35 (M) Be Ig solution. Belg is 100% soluble in aqueous solution. Be IQ (9) = Bezł (q) + 217(69) 1 mole Be Iz dissolues to generate imol of Be 24 and zmol of I ions in the solution. .. - Thus, 0.35(M) Belg contains 0.35(M) Beet iom and (2X0.35) M = 0.7(M) I ions in the solution.

Now, Let'd be the solubility of Pb Iq in 0.35 (M) Be I solution. Ksp for Pb. Iz = [Pb2+] : [I] Pb Ig (3) > Phar) +2 I (29) bixx 2x. In the mixture of 0.35 (M) Be Iq and Pb Ig Solutions Total I ions: come from the 100% dissociation of BeIq . and soluble Pb Iq e. Total [1'] = 0.7(M). I" (from Be In) + 2x (from PbIx) A. Ksp for pb. Iz = x(0.7+ 2x) 2 Therefore, (0.7 +2253 7.1*1ő9 [kSp of Pib ig is constant quantity]:

...org x[(0.7) 7 240.794+ 23)2] = 71x109 org x [ 0.49 + 2.8x + 4x2].2 7.1x109 or, 0.49x + 2.872 + 4x3 = 7.121059 ix is very small : so, sich, a3 term is very very small. Ignoring xh, a3 terms, we get- ::or, 0.49% = 7.1 X 169 22. 7.1x109 . 0.49 sus 1.45x108 a solubility of Pb Iq in 0.35 (M) Be Iq solution becomes 1.45x108 mollut whereas the solubility of Pbiz in Water medium only is 1.21 X163 mollit. Therefore, the solubility of Pb Iz in 0.3.5 (M) Bela solution, decreases comparing to water medium :