Question

<Assignment 13 (Chapter 16) Problem 16.88 ③ 5 of7 Review Constants 1 Periodic Table Consider the titration of 44.0 mL of 0.270 M HF with 0.190 M NaOH. Calculate the pH at each of the following points. Part A How many milliliters of base are required to reach the equivalence point? Express your answer using three significant figures. VNaOH = 62.5 mL Submit Previous Answers Correct Part B Calculate the pH after the addition of 11.0 mL of base Express your answer using two decimal places. IO AL O 2 ? pH = 2.43 Submit Previous Answers Request Answer

B, C, and D are wrong. Please show all work. Thank you! Ka was not provided in the info.

Answer 1

Given that Molarity of HF 0.270 M Volume of HF = 44.0 mL Ka of HF= 3.5 x 104 Concentration of NaOH 0.190 M (a) Number of moles of HF= M*V = 0.270 M* 44.0 mL 11.88 mmol At equivalence point, the no. of moles of NaOH the number of moles of HF Number of moles of NaOH 11.88 mmol Volume of NaOH required for reach equivalence point =moles/molarity 11.88 mmol / 0.190 M 62.53 mL

b) Addition of 10.9 mL of NaOH Number of moles of HF= M*V = 0.270 M* 44.0 mL 11.88 mmol Number of moles of NaOH = M*V = 0.190 M* 11.0 mL 2.09 mmol Net moles of acid 11.88 - 2.09 9.79 mmol Hо HF NaF + NaOH тmol 0 Initial 11.88 0 -2.09 Change -2.09 +2.09 0 9.79 Equilibrium 2.09 According to Henderson equation, [base] PН — pКа+1оg [acid] 2.09 -log (3.5x10log 9.79 pH 3.46 0.67064 = 2.79 pH 2.79

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 62.53 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. 11.88 mmol net moles NaF Total volume (44.0+62.53)mL =0.1115 M 1x10-14 К, of F K KD = 2.86 x 10-1 w 3.5x10 a The dissociation of base is given by HF ОН F +H20 Conc. 0 Initial 0 0.1115 Change - X 0.1115 x X Equilibrium X

х.х =2.86x 10-11 0.1115-x x is small, so 0.1115-x 0.11 15 x2 3.19x1012 x 1.79x10 [OH1.79x10M pOHlog[OH ] =-log (1.79x10)= 5.75 pH 14-5.75 = 8.25 pH 8.25