Determine the mass of the salt formed by the interaction of acetic acid with 20 g of sodium hydroxide

univerkov | February 26, 2021 | education

Problem data: mNaOH is the mass of sodium hydroxide (mNaOH = 20 g).

Const: MNaOH – molar mass of sodium hydroxide (MNaOH ≈ 40 g / mol); MСН3СООNa – molar mass of sodium acetate (anhydrous form, MСН3СООNa ≈ 82 g / mol).

1) The considered reaction: CH3COONa (sodium acetate) + H2O (water) = NaOH (sodium hydroxide) + CH3COOH (acetic acid).

2) The amount of substance taken sodium hydroxide: νNaOH = mNaOH / MNaOH = 20/40 = 0.5 mol.

3) The amount of the substance of the formed sodium acetate: νСН3СООNa / νNaOH = 1/1 and νСН3СООNa = νNaOH = 0.5 mol.

4) The mass of the formed sodium acetate: mСН3СООNa = νСН3СООNa * MСН3СООNa = 0.5 * 82 = 41 g.

Answer: 41 g of sodium acetate should have formed.

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