A 110.0 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