What is the pH of a 0.400M solution of HF

Question

Information for HF

 

Acid: HF 

 

Ka: 7.2×10^-4

 

pKa: 3.35

 

Conjugate Base: NH3

 

Kb: 1.8×10^-5

 

pKb: 4.74

Answer 1

HF <==> H⁺ + F^-

Ka = [H⁺][F^-]/[HF]

7.2x10^-4 = (x)(x)/0.400 

x² = 2.88x10^-4

x = 1.70x10^-2 = [H⁺]

pH = -log [H⁺] 

pH = -log 1.70x10^-2

pH = 1.77

BTW, NH3 is not the conjugate base, F- is, and Kb, pKb and pKa are not relevant. These must be from another question that you received.