Answer:
pH = 3.73 for first part
pH = 3.16 for second part
Explanation:
Use the Henderson-Hasselbalch equation to calculate the pH of this buffer solution:
pH = pKa + log((A⁻)/(HA))
where pka for HF is 3.14
Now we need to calculate the reaction of NaOH with HF in the first part and the reaction of HCl with F⁻ in the second since they are going to change the number of moles present in the buffer after reaction.
mol HF initial = 1.06 mol/L x 0.48 mol/L = 0.51 mol
mol HF reacted with NaOH = 0.16 mol
mol HF final = 0.51 mol - 0.16 = 0.35 mol
mol F⁻ initial = 1.06 L x 1.12 mol/L = 1.19 mol
mol F⁻ final = 1.19 mol + .16 mol = 1.35 mol
pH = 3.14 + log ( 1.35 / 0.35 ) = 3.73
For the addition of HCl we do the same calculations for the reaction of HCl with F⁻:
mol reacted F⁻ = 0.32 mol
mol F⁻ initial = 1.06 L x 1.12 mol/L = 1.19 mol
mol F⁻ final = 1.19 mol - 0.32 0 = 0.87 mol
mol HF final = 0.51 mol + 0.32 mol = 0.83
pH = 3.14 + log ( 0.87/ 0.83 ) = 3.16
