Question

A buffer solution is prepared by adding 0.5 moles of benzoic acid (C6H5COOH, Ka = 6.3 x 10-5) and 0.5 moles of potassium benzoate (C6H5COOK) to enough water to make a 1L solution. How many grams of barium hydroxide is needed to shift the pH of this solution to 6? (This question was previously answered. The answer is not 19.4

Answer 1

Ka of benzoic acid = 6.3 x 10^-5

pKa = 4.2

on addition of C moles of base to acidic buffer. the salt moles increases and acid moles decreases.

pH = pKa + log [salt + C / acid - C]

6 = 4.2 + log [0.5 + C / 0.5 - C]

63.1 = [0.5 + C / 0.5 - C]

C = 0.484

moles of OH- = 0.484

moles of Ba(OH)2 = 0.242

molar mass of Ba(OH)2 = 171.34 g/mol

mass of Ba(OH)2 = moles x molar mass = 0.242 x 171.34

                            = 41.5 g

mass of Ba(OH)2 = 41.5 g