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

| April 29, 2021 | education

Given: w (CH3COOH) = 0.6; mp-pa (NaOH) = 120 g; w (NaOH) = 0.25.

Find: mр-ra (CH3COOH) -?

1. We write down the equation of the proceeding reaction:

CH3COOH + NaOH → CH3COONa + H2O.

2. Find the mass of sodium hydroxide:

m (NaOH) = w (NaOH) * mr-pa (NaOH) = 0.25 * 120 = 30 g.

3. Calculate the chemical amount of sodium hydroxide:

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

M (NaOH) = 23 + 17 = 40 g / mol;

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

4. According to the equation, the acid reacts with alkali in a 1: 1 ratio, set the amount of acetic acid:

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

5. Determine the mass of acetic acid:

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

M (CH3COOH) = 12 + 3 + 12 + 32 + 1 = 60 g / mol;

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

6. Let’s calculate the mass of the acid solution:

mр-pa (CH3COOH) = m (CH3COOH): w (CH3COOH) = 45: 0.6 = 75 g.

Answer: 75 g.



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