Question

Consider the titration of 43.4 mL of 0.265 M HF with 0.195 M NaOH. Calculate the pH at each of the following points.

a. How many milliliters of base are required to reach the equivalence point?

b Calculate the pH after the addition of 10.9 mL of base

c. Calculate the pH at halfway to the equivalence point.

d. Calculate the pH at the equivalence point.

e. Calculate the pH after the addition of 86.8 mL of base.

Answer 1

Given that Molarity of HF 0.265 M Volume of HF = 43.4 mL Ka of HF= 3.5 x 104 Concentration of NaOH 0.195 M (a) Number of moles of HF= M*V = 0.265 M* 43.4 mL 11.501 mmol At equivalence point, the no. of moles of NaOH the number of moles of HF Number of moles of NaOH 11.501 mmol Volume of NaOH required for half equivalence point moles/molarity 11.501 mmol 0.195 M 58.98 mL

b) Addition of 10.9 mL of NaOH Number of moles of HF= M*V = 0.265 M* 43.4 mL 11.501 mmol Number of moles of NaOH = M*V = 0.195 M* 10.9 mL 2.1255 mmol Net moles of acid 11.501 - 2.1255 = 9.3755 mmol Hо HЕ +NaOH NaF тmol 0 Initial 11.501 0 Change -2.1255 -2.1255 +2.1255 0 9.3755 Equilibrium 2.1255 According to Henderson equation, [base] pH %3 pКа +10g [acid] 2.1255 - -log(3.5x 10+log 11 9.3755 pH 3.46 0.64453 2.81 pH 2.81

c) At half equivalence point, the amount of base formed is equal to the amount of acid unreacted So, pH pKa 3.46

d) At equivalence point: Volume of NaOH required to reach equivalence point 58.98 mL This is equivalence point. At this point there is no HF is present in the solution So the pH of the solution is determined by the dissociation of its conjugate base. 11.501 mmol net moles NaF Total volume (43.4+58.98)mL =0.1123 M 1x10-14 К, of F K KD = 2.86 x 10-1 w 3.5x10 a The dissociation of base is given by ОН F HF Conc. +H20 0 0 Initial 0.1123 Change - X 0.1123 x X Equilibrium X

х.х К, 0.1123-x -2.86x10-11 x is small, so 0.1123-x= 0.1123 x23.21x10-12 x=1.79x10 [OH 1.79x10M pOH log[OH ]=-log (1.79x10) = 5.75 pH 14-5.75 = 8.25 pH 8.25 = e) Addition of 86.8 mL of NaOH: Number of moles of HF= M*V = 0.265 M* 43.4 mL 11.501 mmol Number of moles of NaOH M*V = 0.195 M* 86.8 mL 16.926 mmol Number of moles of NaOH present in excess = 37.5 - 16.926 5.425 mmol Total volume of the solution = 43.4 + 86.8 = 130.20 mL Concentration of NaOH = [OH-] 5.425 mmol /130.20 mL 0.041667 M pOH -log[OH-] = -log(0.041667) = 1.38 pOH = 14 - 1.38 = 12.62 pH 14