A110.0 ml sample of 0.20 mhf is titrated with 0.10 mcsoh. determine the ph of the solution after the addition of 440.0 ml of csoh. the ka of hf is 3.5×10−4.

Given:

Concentration of HF = 0.20 m
Volume of HF = 110 ml
Concentration of CsOH = 0.10 m
Volume of CsOH = 440 ml
Ka (HF) = 3.5 x 10^-4

Balanced Chemical Equation:

HF + CsOH ===> CsF + H2O 

pH of the solution = log ([acid]/[base]) + pKa
pH = log ([0.20*(110/1000)]/[0.10*(440/1000)]) - log(3.5x10^-4)
pH = 3.15

Cause its being chemically applied?