Question

How many grams of benzoic acid, C6H5CO2H, are in 3.00 L of a solution whose pH is 2.8? Ka=6.5*10^-5. The answer is 14.7 g benzoic acid but I do not know how they got to that answer. Can someone please show me how to do this question?

Answer 1

we know that

pH = -log [H+]

given

pH = 2.8

so

2.8 = -log [H+]

[H+] = 1.5849 x 10-3

now

benzoic acid is a weak acid

we know that

for weak acids

HA ---> H+ + A-

Ka = [H+] [A-] / [HA]

Ka = [x] [x] / [C-x]

we got

[H+] = [A-] = x = 1.5849 * 10-3

so

6.5 * 10-5 = (1.5849 * 10-3)^2 / ( C - x)

C - x = 0.038644

C = 0.0386444 + x

C = 0.038644 + ( 1.5849 * 10-3)

C = 0.04023

so

the concentration is 0.04023 M

now

moles = molarity x volume (L)

so

moles of benzoic acid = 0.04023 x 3

moles of benzoic acid = 0.12069

now

mass = moles x molar mass

molar mass of benzoic acid = 122.12 g / mol

so

mass of benzoic acid = 0.12069 x 122

mass of benzoic acid = 14.72

so

14.7 grams of benzoic acid is required