Question

4- (a) Calculate the concentration of Cd2+ ion in a solution prepared by mixing 2.0 mL of 1 M CA(NO3)2 solution with 1.0 L of 4.0 M NH3 solution. [Assume that the volume does not change after the addition of 2.0 ml of 1 M Cd(NO3)2] (b) Will you be able to see Ca(OH)2(6) precipitate in the solution? (Kr for Ca(NH3)42+ = 1.0 x 107, Kb for NH3 = 1.8 x 10-5; Ksp for Ca(OH)2 = 5.9 x 10-15 Cd2+ (aq) + 4NH3(aq) –Ca(NH3)+2+ (aq) Ca(OH)2(s) –Cd²+(aq) + 20H(aq)

Answer 1

[Cd (NO3)2 2 las 0.002M [ca2+] = 2X1 1000 [NMg] = 1000x4 1ooo 4 M. Cd²+ (aq) + 4 NH₃ (aq) = {cd (NH) K=107 0.002 M 4 M 4 -0.002 x 4 0.002 M AG totee W ñ 3.992 M ł 0.002 M 0.002 lo7 c (3.99234x2 -6-2= x 7.875810 x 7.875x1513 mare [colet)

NH₃ Canl + H₂O(l) NHy tore, ky a lexis 4 m 4- x n 2 Ч М 23 R 1.8x10-5 4 n²=72815-6 [ope] = x2 2 b . (72x103M Core 8.4858 103M COLLOW, Z Cd2+ (aq) + 2 OH (as) Qa (Cd2+] Compe 7.875 81513 X (8.48581532 2566-96 X 10-19 고 25.67816-17 Anice, Q< (Ksp) =) (No preifitate of colour Сон), formed