Question

W 18. Conside struiaad Consider the titration of a 30.0 mL sample of 0.165 M CSF with 0.300 M HI. The K, of HF is 6.80 x 104. Determine each quantity. [15pts] moles of Esf = 0.65% 0.03 a. The Initial pH. b. The volume required to reach the equivalence point. c. The pH after adding 25 mL of HI. go = 0.0050 M (st: Joti J

Answer 1

Given that concentration of HI = 0.300 M concentration of CsF = 0.165 M. Volume of CSF = 30.0 mL Ka for HF = 6.80 x 10-4 K of F-= 1.0 x 10-14/(6.80 x 10-4) = 1.47 x 10-11 a) Before Addition of any acid: M F + H20 Initial 0.165 Change Equillibrium 0.165 - x K X.X -=1.47x10-11 0.165 - x x is small, so 0.165- X= 0.165 = =1.47x10-11 3x2 = 2.43x10-12 3x=1.56x10-6 :: [OH-] =1.56x100M pOH = -log(1.56x10-)=5.81 pH = 14-5.81= 8.19

Number of moles of F-= M*V = 0.165 M* 30.0 mL = 4.95 mmol At equivalence point, moles of F- = moles of HI Volume of Hl required to reach equivalence point = 4.95 mmol/0.300 M = 16.5 mL At this point, there is no F is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. moles 4.95 mmol [HF]=- -= 0.1065M Total volume (30.0+16.5) mL K, of HF =6.80x104 Dissociation of HF is given by mmol mmol HF HE F Initial 0.1065 Change Equillibrium 0.1065 - x x.x K =- -=6.80x104 0.085 - x x? +6.80x104x–7.24x10-5 = 0 by solving, we get x = 0.008175 :: [H*]=0.008175M pH = -log(0.008175) = 2.09