Question

20

Question 20 Tresma ? Points out of 100 question What is the pH of when 0.073 L of 0.0086 M HCN is titrated to its equivalence point with 0.099 L of NaOH? The Ka of HCN is 4.9x10-10 Multiple tries are permitted; however, 20% (1/5) point will be deducted for each incorrect response.

Answer 1

d 3.34+ 0Co -4 Volume of HON : 0.073 L Molarity of HON : 0.0086 M Molarity of a volume of Nach = 0.0992 Ka of HeN = 4.9810-10 or At equivalence point, the no of moles of base added is equal to the no of moles of acid inihally present in the soluhion Moles of HCN present = 0.073 % x 0.0086 moles - 6.28 x 10" c oles - No of: meles of Nach present # equivalence point a no of moles of acid = 6-28 810-4 moa At equivalence point : No. of moles on (aq) + HEN (aq) H0 (1) + CN- (aq) Before addition 0.00 6 •29 X10-4 0.00 Afande Addition 6029x604 - After addition t_0.00_ 0.00_ - 6•28x10-4 At the equivalence point all the HGN a weak acid (HCN) is ..converted to it.cargugate base (IN) The N" neacts with H2O present in solution and the following equilibrium is set up i n - CN- (aq) + 40 (1) HON (aq) + OH- (aq). Mates of CN at equivalence point : Moles of CN Total volume at equivalence pt. moles of CN- - males af acid + volume of base Moles of Naon at equivalence point 26:28 KL07 wolume of NaOH required to reach equivalence point : D8up 0.0991 concentrahon 6.073 L + 0.099 L = 0.172 L total volume : concentrahon Limples of CN at equivalmcepoint 6.28 X 101. moles 0-1721 3.65 X 10-3 M odes
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CN (aq) + Oh (aa) 0.00 0.00 + x + x Maarity) on CN- (aq) + H₂O (1) Inihal 3.65 X10 -3 - Change on an Equilibrium 1( 3.65X60-3 - x) Guv к Пол Нcy : 4:2 х 10 " ke. K z kw. K & Ko = 2:04 X10 10 x 10-14 4.9 %10-10 ka VA [hen 7 [ou] [cn 1 22. (3.68x10 -3 -2) As n is very small, ko = 3.65 810-3 Or, 2.0481o-S = x2 3.658103 **22 .45 X 10 2.73 x 10-4 x = [on-2 = xis 2.73 x 10-4 Now, ko [4+ ] [on-l -One 1.0X10-14 : 2.9x10 -4 [+] 0.37 X60 3.7 X10 = M pa 2.7x10-4 - log[H+1_ - log (3.7 x 10-1) 2 - Nog 3.7 -11] 10.43