Calculate the mass of 60% acetic acid used to neutralize 120 g of 25% sodium hydroxide solution.

1. Let’s write down the neutralization equation:

NaOH + CH3COOH → CH3COONa + H2O;

2. find the mass of sodium hydroxide:

m (NaOH) = w (NaOH) * m (solution) = 0.25 * 120 = 30 g;

3.determine the chemical amount of hydroxide:

n (NaOH) = m (NaOH): M (NaOH);

M (NaOH) = 23 + 16 + 1 = 40 g / mol;

n (NaOH) = 30:40 = 0.75 mol;

4.Calculate the amount and weight of acetic acid:

n (CH3COOH) = n (NaOH) = 0.75 mol;

m (CH3COOH) = n (CH3COOH) * M (CH3COOH);

M (CH3COOH) = 12 + 3 + 12 + 2 * 16 + 1 = 60 g / mol;

m (CH3COOH) = 0.75 * 60 = 45 g;

5.Calculate the mass of the acid solution:

m (solution) = m (CH3COOH): w (CH3COOH) = 45: 0.6 = 75 g.

Answer: 75 g.