What is the molar solubility of ag2co3 (ksp= 4 × 10–13) in 0.1 m na2co3 solution ?
Answers
ag3co4 (ksp=5×11-14) in 0.1 m na2 cog3 solution
Answer:
molar solubility of is 4.64158 × mol-
molar solubility of in 0.1 m is 1 × mol-
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 = 2x mol-
and = x mol-
so equation is
⇄ 2A +
so
Ksp =
ksp = (2x)² (x)
ksp = 4
put here Ksp value and find x
x³ = ∛(4 × / 4)
x = 4.64158 ×
so here molar solubility of is 4.64158 × mol-
and
we know that 0.1 m na2co3 is added so that
0.1 m is extra ion
so we say
total concentration 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 × = (0.1) × 4
x² = 4 × / (0.1)4
x = √
x = 1 ×
so molar solubility of ag2co3 in 0.1 m na2co3 is 1 × mol-