Question

Calculate the pH at the equivalence point in titrating 0.048 M solutions of each of the following with 0.016 M NaOH.
(a) hydroiodic acid (HI)
(b) hydrosulfuric acid (H₂S), K_a = 9.5e-08
(c) phenol (HC₆H₅O), K_a = 1.3e-10

Answer 1

(A) Hland NaOH are strong acid and base, So, at equivalence point, pH = 7 (B) Conc. of H2S = 0.048 M. Let the V be the volume of H2S solution. Number of moles of H2S = 0.048 M * V mL = 0.048V At equivalence point, number of moles of NaOH = moles of H2S = 0.048 V Volume of NaOH required to reach equivalence point = 0.048V /0.016 = 3V

At this point, there is no H S is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. [HS]== moles 0.048V mmol = 0.012M Total volume (V +3V) mL K, of HS- = 1x10-14 9.5x10-8 =1.02x10-1 Dissociation of HF is given by Ka mmol HS + H20 H2S + OH Initial 0.012 Change Equillibrium 0.012 - x x.x -= 1.02 x 10-7 0 0.012 - x x2 +1.02x10-ºr-1.22x10-9 = 0 by solving, we get x = 3.49x10-5 :: [OH]=3.49x10-M pOH = -log(3.49x10) = 4.46 pH = 14 - pOH = 14 – 4.46 = 9.54

(C) Conc. of CH3OH = 0.048 M. Let the V be the volume of CH3OH solution. Number of moles of CH,OH = 0.048 M * V mL = 0.048V At equivalence point, number of moles of NaOH = moles of CH3OH = 0.048 V Volume of NaOH required to reach equivalence point = 0.048V /0.016 = 3V

At this point, there is no CGH,OH is present in the solution. So the pH of the solution is determined by the dissociation of its conjugate acid. 0.048V mmol [CH,0"]= -=0.012M Total volume (V + 3V ) mL moles Ky of CH,0 K; of CH,0 = 75x10-8 - 1x10-14 0-8 = 7.69x10-5 Dissociation of CGH,O is given by 1 Ka mmol C6H50+ H20 CGH OH + OH Initial 0.012 Change Equillibrium 0.012 - X K. _ X.X KB 0.012-x -= 7.69 x10-5 x² + 7.69x10-x-9.23x10-7 = 0 by solving, we get x=0.000923 ::[OH-]=0.000923M pOH = -log(0.000923)=3.03 pH = 14 – POH = 14 – 3.03 = 10.97