Question

Calculate the pH of the solution that results from the following mixture.

140.0 mL of 0.24 M HF with 230.0 mL of 0.32 M NaF Express your answer using two decimal places.

Answer 1

Ka of HF = 6.6*10^-4

Concentration after mixing = mol of component / (total volume)

M(NaF) after mixing = M(NaF)*V(NaF)/(total volume)

M(NaF) after mixing = 0.32 M*230.0 mL/(230.0+140.0)mL

M(NaF) after mixing = 0.1989 M

Concentration after mixing = mol of component / (total volume)

M(HF) after mixing = M(HF)*V(HF)/(total volume)

M(HF) after mixing = 0.24 M*140.0 mL/(140.0+230.0)mL

M(HF) after mixing = 9.081*10^-2 M

Ka = 6.6*10^-4

pKa = - log (Ka)

= - log(6.6*10^-4)

= 3.18

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 3.18+ log {0.1989/9.081*10^-2}

= 3.521

Answer: 3.52