Help ya 1. A 250 cm3 of solution contains 132 mg of ammonium sulphate, (NH4) 2SO4 If dissociation constant of ammonia is 2.0 x 10¯5 what is the pH of the soluti
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1. A 250 cm3 of solution contains 132 mg of ammonium sulphate, (NH4) 2SO4 If dissociation constant of ammonia is 2.0 x 10¯5 what is the pH of the solution? The molecular mass of (NH4) 2SO4 id 132 /mol)
2 . one litre of sodium acetate solution has a pH of 9.0 Ka for CH3COOH is 10¯5 how many grams of sodium acetate present in solution? The molecular mass of CH3COONa is 82
1. A 250 cm3 of solution contains 132 mg of ammonium sulphate, (NH4) 2SO4 If dissociation constant of ammonia is 2.0 x 10¯5 what is the pH of the solution? The molecular mass of (NH4) 2SO4 id 132 /mol)
2 . one litre of sodium acetate solution has a pH of 9.0 Ka for CH3COOH is 10¯5 how many grams of sodium acetate present in solution? The molecular mass of CH3COONa is 82
1 Jawaban
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1. Jawaban sandysuhendra
1) mole of (NH4)2SO4 = 132/132 = 1 mmole
[(NH4)2SO4] = M = 1/250 = 4.10⁻³ M
[H+] = √(Kb. M. valency) = √(2.10⁻⁵×4.10⁻³×2) = 4.10⁻⁴ M
pH = - log 4.10⁻⁴ = 4 - log 4 = 4 - 2 log 2 = 3,4
2) pH=9 —> pOH=5 —> [OH-] = 10⁻⁵ M
[OH-] = √(Ka. M. valency)
10⁻⁵ = √(10⁻⁵×M×1)
10⁻¹⁰ = 10⁻⁵. M
M = 10⁻⁵ M
mole of CH3COONa = V. M = 1×10⁻⁵ mole
Mass of CH3COONa = 10⁻⁵ × 82 = 0,00082 grams