Question

What is the molar solubility of ag2co3 (ksp= 4 × 10–13) in 0.1 m na2co3 solution ?

Answer 1

Molar solubility:- 
ag3co4 (ksp=5×11-14) in 0.1 m na2 cog3 solution

Answer 2

Answer:

molar solubility of $Ag_{2}CO_{3}$ is 4.64158  × $10^{-5}$ mol-$L^{-1}$

molar solubility of  $Ag_{2}CO_{3}$ in 0.1 m  $Na_{2}CO_{3}$ is 1 × $10^{-6}$ mol-$L^{-1}$

Explanation:

Given data

ksp = 4 × 10^–13

na2co3 solution = 0.1 m

to find out

molar solubility of ag2co3 in 0.1 m na2co3

solution

we consider here solubility of ag2co3 is x

so A$g^{+}$ = 2x mol-$L^{-1}$

and $CO^{-2}_3$ = x mol-$L^{-1}$

so equation is

$Ag_{2}CO_{3}$ ⇄ 2A$g^{+}$ +  $CO^{-2}_3$

so

Ksp  = $A(g^{+})^{2}$ $CO^{-2}_3$

ksp = (2x)² (x)

ksp = 4

put here Ksp value and find x

x³ = ∛(4 × $10^{-13}$ / 4)

x = 4.64158  × $10^{-5}$

so here molar solubility of $Ag_{2}CO_{3}$ is 4.64158  × $10^{-5}$ mol-$L^{-1}$

and

we know that 0.1 m na2co3 is added so that

0.1 m $CO^{-2}_3$  is extra ion

so we say

total concentration $CO^{-2}_3$  is = x + 0.1

so that

Ksp = 2x² ( x+0.1)

Ksp  =  (0.1) × 4x²

and we know x is very small as 0.1 so we neglect it and put Ksp value here

4 × $10^{-13}$ = (0.1) × 4

x² = 4 × $10^{-13}$ / (0.1)4

x = √$10^{-12}$

x = 1 × $10^{-6}$

so molar solubility of ag2co3 in 0.1 m na2co3 is 1 × $10^{-6}$ mol-$L^{-1}$