Question

What mass of sodium benzoate should be added to 150.0 mL of a 0.17 M benzoic acid solution in order to obtain a buffer with a pH of 4.30? (answer is in grams).

Answer 1

Ka of benzoic acid = 6.3*10^-5

Ka = 6.3*10^-5

pKa = - log (Ka)

= - log(6.3*10^-5)

= 4.2007

use formula for buffer

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

4.3 = 4.2007 + log ([C6H5COONa]/[C6H5COOH])

log ([C6H5COONa]/[C6H5COOH]) = 0.0993

[C6H5COONa]/[C6H5COOH] = 1.257

[C6H5COONa] = 0.2137

volume , V = 150.0 mL

= 0.15 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.2137*0.15

= 3.205*10^-2 mol

Molar mass of C6H5COONa = 7*MM(C) + 5*MM(H) + 2*MM(O) + 1*MM(Na)

= 7*12.01 + 5*1.008 + 2*16.0 + 1*22.99

= 144.1 g/mol

we have below equation to be used:

mass of C6H5COONa,

m = number of mol * molar mass

= 3.205*10^-2 mol * 144.1 g/mol

= 4.62 g

Answer: 4.62 g