Question

Calculate the pH after 0.16 mole of NaOH is added to 1.06 L of a solution that is 0.48 M HF and 1.12 M NaF, and calculate the pH after 0.32 mole of HCl is added to 1.06 L of the same solution of HF and NaF.

Answer 1

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