Question

1. How many grams of sodium lactate, NaC3H5O3, should be added to 1.00 L of 0.150 M lactic acid, HC3H5O3, to form a buffer solution with a pH of 3.90? (Ka = 1.4*10^-4 for lactic acid)

How do you do this?

Answer 1

Ka = 1.4*10^-4

pKa = - log (Ka)

= - log(1.4*10^-4)

= 3.8539

use formula for buffer

pH = pKa + log ([NaC3H5O3]/[HC3H5O3])

3.9 = 3.8539 + log ([NaC3H5O3]/[HC3H5O3])

log ([NaC3H5O3]/[HC3H5O3]) = 0.0461

[NaC3H5O3]/[HC3H5O3] = 1.1121

[NaC3H5O3] = 0.1668

volume , V = 1.0 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.1668*1

= 0.1668 mol

Molar mass of NaC3H5O3 = 1*MM(Na) + 3*MM(C) + 5*MM(H) + 3*MM(O)

= 1*22.99 + 3*12.01 + 5*1.008 + 3*16.0

= 112.06 g/mol

we have below equation to be used:

mass of NaC3H5O3,

m = number of mol * molar mass

= 0.1668 mol * 112.06 g/mol

= 18.7 g

Answer: 18.7 g