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