chemistry ?

1.

Refer: https://owl.oit.umass.edu/departments/Chemistry/appendix/ksp.html

Ksp(MnCO₃) = 1.8 × 10⁻¹¹
(Ksp values from different sources may be slightly different.)

(NH₄)₂CO₃ is soluble in water, and each mole of (NH₄)₂CO₃ gives 1 mole of CO₃²⁻ ions.

______ MnCO₃(s) ⇌ Mn²⁺(aq) + CO₃²⁻(aq) ___ Ksp = 1.8 × 10⁻¹¹
Initial (M): __________ 0 ______ 0.178
Change (M) ________ +s _______ +s
Equilibrium (M): _____ s ______(0.178 + s)

As Ksp is very small and due to the common ion effect in the presence of CO₃²⁻ ions, the solubility is very low.
It can be assumed that 0.178 ≫ s
Equilibrium [CO₃²⁻] = (0.178 + s) M ≈ 0.178 M

At equilibrium:
Ksp = [Mn²⁺] [CO₃²⁻]
1.8 × 10⁻¹¹ = s × 0.178
s = (1.8 × 10⁻¹¹) / 0.178 = 1.01 × 10⁻¹⁰

Max. amount of MnCO₃ that will dissolve = 1.01 × 10⁻¹⁰ M

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2.
Refer: https://owl.oit.umass.edu/departments/Chemistry/appendix/ksp.html
Ksp(Ni(CN)₂) = 3.0 × 10⁻²³
(Ksp values from different sources may be slightly different.)

Ni(CH₃COO)₂ is soluble in water, and each mole of Ni(CH₃COO)₂ gives 1 mole of Ni²⁺ ions.

______ Ni(CN)₂(s) ⇌ Ni²⁺(aq) + 2CN⁻(aq) ___ Ksp = 3.0 × 10⁻²³
Initial (M): __________ 0.260 ____ 0
Change (M) __________ +s ____ +2s
Equilibrium (M): ___ (0.260 + s) __ 2s

As Ksp is very small and due to the common ion effect in the presence of Ni²⁺ ions, the solubility is very low.
It can be assumed that 0.260 ≫ s
Equilibrium [Ni²⁺] = (0.260 + s) M ≈ 0.260 M

At equilibrium:
Ksp = [Ni²⁺] [CN⁻]²
3.0 × 10⁻²³ = 0.260 (2s)²
s = √[(3.0 × 10⁻²³) / (0.260 × 4)] = 5.37 × 10⁻¹²

The molar solubility of Ni(CN)₂ = 5.37 × 10⁻¹² M