Question

17. The acid/base titration curve below has been obtained by titrating 0.10 moles of a weak acid of acidity constant Ka=8.0x10-5 in 100.0 mL of water with a strong base (NaOH). 8pts 0 moles NaOHH Answer the following questions (neglect dilution): (a) The equivalence point C has been reached after adding moles of NaOH. (b) Compute the pH at the start of the titration (point A): (c) Compute the pH at the midpoint of the titration (point B): (d) Compute the pH at the equivalence point (point C):

Show calculations please. Thank you!

Answer 1

The balance equation for the reaction between weak acid and NaOH is HA + NaOH- NaA H2O Number of moles of HA 0.10 moles Volume of water 100 mL a) According to balanced equation, of acid At equivalence point, the number of moles of NaOH number of moles So number of moles of NaOH to be added to reach equivalence point 0.10 mol bl Number of moles of HA 0,10 mol volume of HA 100 mL Conc. of HA = moles/volume = 0.10 mol/0.1 L = 1 M НА Initial 1.0 Change Equilibrium

a-8.0x10-5 x2 +8.0× 10 5x-8.0x10 5-0 by Solving, x =-0.00898 and 0.00890 x cannot be negative, so x = 0.0089 H 0.0089M pH = _ log[T ] =-log(0.0089) = 2.05 c) At mid point number of moles of NaOH 2(Number of moles of HA) At this point, conc. of acid and its conjugate base is same HA H,O Initial 0.1 0 0 Change 0.5 0.5 0.5 Equilibrium 0.5 0 0.5 net moles0.5 mol HAT [NaA]= According to Henderson equation, Total volume V net moles 0.5 mol Total volumeV 0.5 mol pH = pKa + log --log(8.0x10)+logl 0.5 mol pH=41+0 plI 4.1

d) At equivalence point: Let the conc. of NaOH be1 M Number of moles of HA = 0.1 mol At equivalence point the no. of moles of NaOH-the number of moles of HA Volume of NaOH required for equivalence poinft 0,1 mol moles/molarity = 0.1 mol/ 1M = 100 mL 0.1mol0.1mol 0.1mol Total volume 100+100 200mL 0.2 L [A] moles 0.5 M K, K, 1x10ー=1.25x10-10 8x10-5 The dissociation of base is given by concA +H2O HA он Initial 0.5 0 0 Change + X + X Equilibrium 0.5 X 1.25 × 10 10 b0.5-x x is small, so 0.5-x-0.5 x' = 0.5 × 1.25 × 10-10 x-7.9x106 [OH-7.9 × 106M nQH--log OH-_-log(7.9 × 10-6)-5.1 pH 14 pOH 14-5.1-8.9