Question

13. The value of Rsp for AgCl is 1.8x10-10 What is the solubility (mol/L) of AgCl in 0.13M NaCl(aq)? How many mols of AgCl would dissolve in 1.00L of 0.13M NaCl(aq) at 25°C?

Answer 1

Consider a reaction ,AgCl (s)  $\rightleftharpoons$ Ag ⁺ (aq) + Cl ^- (aq)

For above reaction,equilibrium constant  is K sp = [ Ag ⁺ ] [ Cl ^- ] = 1.8 $\times$ 10 ^-10

If Y is the solubility of AgCl in mol / L , then [ Ag ⁺ ] = Y mol / L and [ Cl ^- ] = (Y + 0.13 ) mol / L.

$\therefore$  K sp = [ Ag ⁺ ] [ Cl ^- ]

K sp = Y $\times$ (Y + 0.13 ) = 1.8 $\times$ 10 ^-10

Assume Y is very small as compared to 0.13, hence we can write Y + 0.13 $\approx$   0.13

$\therefore$ Y $\times$   0.13 = 1.8 $\times$ 10 ^-10

Y = 1.8 $\times$ 10 ^-10 / 0.13

Y =  1.38 $\times$ 10 ^-09 M

ANSWER : Solubility of AgCl in 0.13 M NaCl solution = 1.4 $\times$ 10 ^-09 mol / L.

1.4 $\times$ 10 ^-09 moles of AgCl will dissolve in 0.13 M NaCl solution.