Let s = the molar solubility.
Ag₂CO₃(s) ⇌ 2Ag⁺(aq) + CO₃²⁻(aq); = 8.10 × 10⁻¹²
E/mol·L⁻¹: 2s s
=[Ag⁺]²[CO₃²⁻] = (2s)²×s = 4s^3 = 8.10 × 10⁻¹²
write the equation for dissociation of AgCO3
Ag2CO3 > 2Ag^+ + CO3^2-
let the concentration of co3^-2 be represented by X , this implies that the concentration of Ag^+ = 2x
8.10 x10^-12 is therefore= (2x^2)(x)
=8.10 x10^-12 = 4 x^3, divide both side by 4
=2.025 x10^-12 =x^3, find the cube root
solubity in grams per liter is therefore=0.000127 x molar mass of Ag2CO3(275.75g/mol)
that is 0.000127 x275.75 =0.035 g/l
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