An excess of copper (II) sulfate solution was added to the sodium hydroxide solution. This resulted in the formation

An excess of copper (II) sulfate solution was added to the sodium hydroxide solution. This resulted in the formation of a precipitate weighing 4.9 g. Determine the mass of the initial alkali solution.

Given:
ω (NaOH) = 5%
m (sediment) = 4.9 g

To find:
m solution (NaOH) -?

Solution:
1) 2NaOH + CuSO4 => Cu (OH) 2 ↓ + Na2SO4;
2) M (Cu (OH) 2) = Mr (Cu (OH) 2) = Ar (Cu) + Ar (O) * 2 + Ar (H) * 2 = 64 + 16 * 2 + 1 * 2 = 98 g / mol;
M (NaOH) = Mr (NaOH) = Ar (Na) + Ar (O) + Ar (H) = 23 + 16 + 1 = 40 g / mol;
3) n (Cu (OH) 2) = m (Cu (OH) 2) / M (Cu (OH) 2) = 4.9 / 98 = 0.05 mol;
4) n (NaOH) = n (Cu (OH) 2) * 2 = 0.05 * 2 = 0.1 mol;
5) m (NaOH) = n (NaOH) * M (NaOH) = 0.1 * 40 = 4 g;
6) m solution (NaOH) = m (NaOH) * 100% / ω (NaOH) = 4 * 100% / 5% = 80 g.

Answer: The mass of the NaOH solution is 80 g.