Question

Question 1 : The solubility product, Kp. of Cd3(PO4)2 is 2.5 x 10-33 What is the solubility (in g/L) of Cd3(PO4), in pure water? Question 2 The solubility product of Cu(OH), is 4.8 x 10-20 Calculate the value of pCu2+ or -log(Cu2+1, in an aqueous solution of NaOH which has a pH of 12.84 and is saturated in Cu(OH)2 Question 3 The equilibrium constant for the formation of Cu(CN), is 2.0 x 1030 Calculate the value of pCu2+ or -log(Cu2+1, if we were to dissolve 2.35 g of CuCl, in 1.000 L of a solution 0.890 M in NaCN. The addition of CuCl2 does not alter the volume (the final volume is still 1.000 L).

Answer 1

Answer:

Question 1.

Solubility of Cd3(PO4) 2 in water is 1.1828*10^-7 g/L in water.

Explaination:

y+ 2 - ksp=[Art] ². [ 3²-] Relation b/w soleability and soleability product Ax By ř x + y B ys leqm initial moles s as [ Att as [B²] =ys こ (15)*(45) y こ : Ksp aty z yo S

3 ed+2 -3 + 2 POL Auswen 17 Cd₂ (PO4)2 ksp = (x)? sx+y (3+2) = 1333 (29². s الدو "/5 1/5 1/5 -35 ... S= ksp 108 108 (2.314814815x16 = 1.1828x10-7 Agth 2,5x10-33

Answer 2

0.0000624 g/L