What mass of Ca(OH)2 is required to react with the acetic acid (CH3CO2H) in 25.0 mL of a solution having a density of 1.065 g/mL and containing 58.0% acetic acid by mass?

9.53g

Explanation:

Given the following :

Density = 1.065g/mL

Volume of solution = 25.0mL

% of acetic acid =58%

Molar mass of CH3COOH = 60.05g/mol

Molar mass of Ca(OH)2 = 74.093g/mol

Reaction:

2CH3COOH + Ca(OH)2 > Ca(CH3Co2)2 +2H2O

Volume of solution × density × % of CH3COOH × (mole per molar mass CH3COOH) × (mole of Ca(OH)2 per mole of CH3COOH) × (molar mass of Ca(OH)2 per mole of Ca(OH)2)

25mL × (1.065/1) × (58/100) × (1 / 60.05) × (1 / 2) × (74.093 / 1)

25 × 1.065 × 0.58 × 0.0166527 × 0.5 × 74.093

= 9.526g

the ratio of the conjugate base to the acid required to attain the desired ph.