An excess of copper (II) sulfate solution was added to a solution with a mass fraction of sodium hydroxide of 15%. This formed a precipitate weighing 24.5 g. Determine the mass of the initial alkali solution.
1) Let’s compose the reaction equation:
2NaOH + CuSO4 = Cu (OH) 2 ↓ + Na2SO4
2) Calculate the amount of the precipitated copper (II) hydroxide:
n (Cu (OH) 2) = m (Cu (OH) 2) / M (Cu (OH) 2) = 24.5 / 98 = 0.25 mol.
3) Determine the amount of the substance, the mass of sodium hydroxide, the mass of the initial solution:
according to the reaction equation
n (NaOH) = 2n (Cu (OH) 2) = 2 ∙ 0.25 = 0.5 mol
m (NaOH) = n (NaOH) ∙ M (NaOH) = 0.5 ∙ 40 = 20 g.
m (solution) = m (NaOH) ∙ 100% / ω (NaOH) = 20 ∙ 100/15 = 133.33 g.
Answer: 133.33 g.
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