Calculate the pH of a solution that is 0.40 M HF and 1.00 M KF.?

Refer to: http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf
Ka for HF = 7.2 × 10⁻⁴ and pKa for HF = 3.14
(Different pKa values would lead to different answers.)

Consider the dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq)

pH = pKa + log([F⁻]/[HF]) = 3.14 + log(1.00/0.40) = 3.54