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Question:

If Ksp of Ag2CO3 is 8 × 10⁻¹², the molar solubility of Ag2CO3 in 0.1 M AgNO3 is:

8 × 10⁻³M

8 × 10⁻²M

8 × 10⁻¹M

8 × 10⁻⁰M

Solution:

Ag2CO3(s)⇌2Ag+(aq)(0.1+2S)M + CO₃⁻²(aq)SM
Ksp=[Ag+]²[CO₃⁻²]
8 × 10⁻¹²=(0.1+2S)²(S)
As Ksp is so small, so 0.1+2S=0.1
S=8 × 10⁻¹⁰M.