Question

What is the pH of the solution if 0.021 g of sodium fluoride is added to 130.0 mL of a 0.098 M HF solution? Ka for HF is 6.8

Answer 1

Molar mass of NaF,

MM = 1*MM(Na) + 1*MM(F)

= 1*22.99 + 1*19.0

= 41.99 g/mol

mass(NaF)= 0.021 g

use:

number of mol of NaF,

n = mass of NaF/molar mass of NaF

=(2.1*10^-2 g)/(41.99 g/mol)

= 5.001*10^-4 mol

volume , V = 1.3*10^2 mL

= 0.13 L

use:

Molarity,

M = number of mol / volume in L

= 5.001*10^-4/0.13

= 3.847*10^-3 M

Given:

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

use:

pH = pKa + log {[conjugate base]/[acid]}

= 3.167+ log {3.847*10^-3/9.8*10^-2}

= 1.76