Question

What mass of sodium benzoate should be added to 160.0mL of a 0.16M benzoic acid solution in order to obtain a buffer with a pH of 4.25?

Express your answer using two significant figures.

Answer 1

Ka of benzoic acid = 6.3*10^-5

Ka = 6.3*10^-5

pKa = - log (Ka)

= - log(6.3*10^-5)

= 4.201

use formula for buffer

pH = pKa + log ([C6H5COOH]/[C6H5COONa])

4.25 = 4.2007 + log ([C6H5COOH]/[C6H5COONa])

log ([C6H5COOH]/[C6H5COONa]) = 0.0493

[C6H5COOH]/0.16 = 1.1203

[C6H5COOH] = 0.1793

volume , V = 1.6*10^2 mL

= 0.16 L

use:

number of mol,

n = Molarity * Volume

= 0.1793*0.16

= 2.868*10^-2 mol

Molar mass of C6H5COOH,

MM = 7*MM(C) + 6*MM(H) + 2*MM(O)

= 7*12.01 + 6*1.008 + 2*16.0

= 122.118 g/mol

use:

mass of C6H5COOH,

m = number of mol * molar mass

= 2.868*10^-2 mol * 1.221*10^2 g/mol

= 3.502 g

Answer: 3.5 g