Tentukan pH Mg(OH)2 jika ke dalam larutan MgCl2 0,2M ditambahkan NH4OH sehingga

terbentuk endapan Mg(OH)2 yang memiliki harga Ksp = 2x10-11!

pH 8.89

Explanation:

English Translation

If the MgCl₂ solution of 0.2 M has its pH raised by adding NH₄OH, the precipitate will begin to form at a pH of approximately.

Given the solubility product (Ksp) of Mg(OH)₂ = 1.2 x 10⁻¹¹

Assuming all of the salts involved all ionize completely

MgCl₂ ionizes to give Mg²⁺ and Cl⁻

MgCl₂ ⇌ Mg²⁺ + 2Cl⁻

1 mole of MgCl₂ gives 1 moles of Mg²⁺

Since the concentration of Mg²⁺ is the same as that of MgCl₂ = 0.2 M

Mg(OH)₂ is formed from 1 stoichiometric mole of Mg²⁺ and 2 stoichiometric moles of OH⁻

Ksp Mg(OH)₂ = [Mg²⁺][OH⁻]²

(1.2 x 10⁻¹¹) = 0.2 × [OH⁻]²

[OH⁻]² = (6×10⁻¹¹)

[OH⁻] = √(6×10⁻¹¹)

[OH⁻] = 0.000007746 M

p(OH) = - log [OH⁻] = - log (0.000007746)

pOH = 5.11

pH + pOH = 14

pH = 14 - pOH = 14 - 5.11 = 8.89

Hope this Helps!!!

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it's filtration, but this is an assumption so my apologies if i got this