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babunello [35]
3 years ago
9

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

Chemistry
1 answer:
Vadim26 [7]3 years ago
3 0

Answer:

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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The number of C atoms in 0.524 moles of C is 3.15 atoms.

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A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

A. The number of C atoms in 0.524 mole of C:

6.02214076 × 10^{23} x 0.524 mole

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Learn more about moles here:

brainly.com/question/8455949

#SPJ1

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