Question

How many grams of solid sodium fluoride should be added to 0.500 L of a 0.248 M hydrofluoric acid solution to prepare a buffer with a pH of 3.988 ? grams sodium fluoride = g.

Answer 1

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.4559

use formula for buffer

pH = pKa + log ([NaF]/[HF])

3.988 = 3.4559 + log ([NaF]/[HF])

log ([NaF]/[HF]) = 0.5321

[NaF]/[HF] = 3.4046

[NaF] = 0.8443

volume , V = 0.5 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.8443*0.5

= 0.4222 mol

Molar mass of NaF = 1*MM(Na) + 1*MM(F)

= 1*22.99 + 1*19.0

= 41.99 g/mol

we have below equation to be used:

mass of NaF,

m = number of mol * molar mass

= 0.4222 mol * 41.99 g/mol

= 17.7 g

Answer: 17.7 g