Question

What is the pH of a buffer that is prepared by mixing 30.0 mL of 1.0 M HF and 20.0 mL of 2.0 M KF? (K a a for HF 7.2 x 10 −4 )

Answer 1

HF is a weak acid and KF is a salt which dissociates to give F^-

which is the conjugate base of HF.

Also 1 L=1000 mL

As 1 mol KF gives 1 mol F^- in solution

So Number of moles of F^{​​​​​​-}=Numbe of moles of KF=Molarity x Volume =2.0 M x 20.0 mL/1000 mL/L=0.040 mol

Number of moles of HF=Molarity x Volume (L)=1.0 M x 30.0 mL/1000 mL/L

=0.030 mol

Total volume of the buffer=20.0 mL+30.0 mL=50.0 mL/1000 mL/L=0.050 L

By Henderson-Hasselbalch equation, pH of an acidic buffer is given as

Where Ka=acid dissociation constant for HF=7.2x10^-4

pKa=-logKa=-log(7.2x10^-4)=3.14

[A^{​​​​​-}]=Concentration of conjugate base=concentration of F^-=Number of moles of F^{​​​​​-}/volume of the solution=0.040 mol/0.050 L

[HA]=concentration of weak acid=concentration of HF=Number of moles of HF/volume of the solution

=0.030 mol/0.050 L

Substituting the values in Henderson-hasselbalch equation

pH=

=

=3.14+0.12=3.26

So pH of given buffer=3.26