103.0 mL of a solution of quinine contains 144.2×10 3 mol of quinine. This is ti

Question

103.0 mL of a solution of quinine contains 144.2×103 mol of quinine.
This is titrated with 0.200 M HCl, and the pH half-way to thestoichiometric point is 8.90. Quinine is monobasic with Kb = 7.90 ×106. What is the pH at the stoichiometric point for the titrationin question 6 ? Quinine is monobasic with Kb = 7.90 ×106. What is the pH at the stoichiometric point for the titrationin question 6 ?

Explanation / Answer

quinine reacts with HCl in 1:1 ratio So need 144.2e-3 /2 mol HCl 144.2e-3 mol HCl /2* ( 1L / 0.2 mol) = 0.3605 L added Final volume = 360.5 mL + 103.0 mL = 363.5 mL 144.2 e-3 /2 mol of protonaed quinine also formed B + H+   =>  BH+             (B is quinine and BH+ is protonated) [B] = [BH+] = 144.2e-3/2 mol / 0.363 L = 0.198349381 M             BH+ +    H2O      B  +          H3O+ init         0.198                         0.198           change   -x                                   x               + x final      0.198 -x                       0.198 + x        + x Ka = Kw/Kb = 1e-14/Kb = 1e-14/7.9e-6 = 1.26582278e-9 Ka = [x][0.198+x]/(0.198 - x) = 1.266 e-9 Assume that 0.198 + x ~ 0.198    and 0.198 -x ~0.198 x = 1.266e-9 = [H3O+] pH = -log[H3O+] = 8.90 [H+] = 1e-7 - 9.87e-8 =

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