How many grams of AgSCN(Ksp=5*10^-12) dissolve in a) 0,005M KSCN b) 2,3*10^-5 M AgNO3?

Molar mass of AgSCN = (107.9 + 32.1 + 12.0 + 14.0) g/mol = 166.0 g/mol

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In 0.05 M KSCN:
_________ AgSCN(s) _____ ⇌ _____ Ag⁺(aq) _____ + SCN⁻(aq) _____ Ksp = 5 × 10⁻¹²
Initial: __________________________ 0 M _________ 0.005 M
Change: ________________________ +s M _________ + s M
Equilibrium: _____________________ s M _______ (0.005 + s) M

As Ksp for AgSCN is very small and due to the common ion effect in the presence of SCN⁻,
[SCN⁻] due to the dissociation of AgSCN is negligible.
Hence, [SCN⁻] at equilibrium = (0.005 + s) M ≈ 0.005 M

At equilibrium: Ksp = [Ag⁺] [SCN⁻]
5 × 10⁻¹² = s × 0.005
s = 1 × 10⁻⁹ M

Mass of AgSCN dissolves in 1 L of 0.005 M KSCN = (1 × 10⁻⁹ mol/L) × (1 L) × (166.0 g/mol) = 1.66 × 10⁻⁷ g
Mass of AgSCN dissolves in 0.005 M KSCN = (1.66 × 10⁻⁷ g/L) × (Volume of the solution in L)

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In 2.3 × 10⁻⁵ M AgNO₃ :
_________ AgSCN(s) _____ ⇌ _____ Ag⁺(aq) _____ + SCN⁻(aq) _____ Ksp = 5 × 10⁻¹²
Initial: _______________________ 2.3×10⁻⁵ M ______ 0 M
Change: ________________________ +s M ________ + s M
Equilibrium: ________________ (2.3×10⁻⁵ + s) M ___ s M

At equilibrium: Ksp = [Ag⁺] [SCN⁻]
5 × 10⁻¹² = (2.3 × 10⁻⁵ + s) × s
s² + (2.3 × 10⁻⁵)s - (5 × 10⁻¹²) = 0
s = {-(2.3 × 10⁻⁵) ± √[(2.3 × 10⁻⁵)² + 4×(5 × 10⁻¹²)]} / 2
s = 2.15 × 10⁻⁷ or s = -2.32 × 10⁻⁷ (rejected)

Mass of AgSCN dissolves in 1 L of 2.3 × 10⁻⁵ M AgNO₃ = (2.15 × 10⁻⁷ mol/L) × (1 L) × (166.0 g/mol) = 3.57 × 10⁻⁵ g
Mass of AgSCN dissolves in 2.3 × 10⁻⁵ M AgNO₃ = (3.57 × 10⁻⁵ g/L) × (Volume of the solution in L)