A 100.0 mL sample of 0.20 M HF is titrated with 0.10 M KOH. Determine the pH of the solution after the addition of 75.0 mL of KOH. The Ka of HF is 3.5 × 10-4

The pH of the solution is 3.23

Explanation:

The reaction between HF and KOH is:

KOH + HF → KF + H₂O

The number of moles of HF in 100.0 mL is:

1000.0 mL 0.20 mol

100.0 mL x= 0.02 mol

The number of moles of KOH in 75.0 mL is:

1000.0 mL 0.10 mol

75.0 mL x= 7.5×10⁻³ mol

We work with moles because when two solutions are mixed, the  solution volume increases. As the volume increases, molarity will change but the  number of moles will remain the same.

The changes in the number of moles are:

                               KOH      +      HF      →        KF         +      H₂O

Initial (mol)            7.5×10⁻³        2.0×10⁻²          0                  

Change (mol)       -7.5×10⁻³      -7.5×10⁻³        7.5×10⁻³    

Final (mol)                  0             1.25×10⁻²       7.5×10⁻³    

At this stage we have a buffer system made up of HF and F⁻ (from  the salt, KF). To calculate the pH of the solution, we write:

Ka= [H⁺][F⁻] / [HF]

[H⁺]=  [HF]Ka / [F⁻] = 5.83×10⁻⁴

Therefore,    pH= 3.23