Question

Calculate the pH at the equivalence point in titrating 0.050 M solutions of each of the following with 0.043 M NaOH.




(a) hydrochloric acid (HCl)

pH =  



(b) ascorbic acid (HC₆H₇O₆), K_a = 8e-05

pH =  



(c) hypoiodous acid (HIO), K_a = 2.3e-11

pH =  

Answer 1

a) HCl is a strong acid and NaOH is a strong base.
At equivalence point, the pH =7.

(b) At equivalence point, mmoles of acid equal mmoles of base.
That is,

Mbase x Vbase = Macid x Vacid

thus Vacid = (Mbase x Vbase) / Macid

What we need to solve this problem is the total volume at the equivalence point and the number of mmoles of the salt that is formed:

[C6H7O6-] = (Mbase x Vbase )/(Vbase + Vacid) = (Mbase x Vbase) / (Vbase + (Mbase x Vbase) / Macid)

[C6H7O6-] = Mbase /(1 + (Mbase/Macid)) = MbaseMacid/(Mbase+Macid)

[C6H7O6-] = (0.043)(0.050)/(0.043+0.050) = 0.0231 M

C6H7O6-(aq) + H2O(l) <====> HC6H7O6 (aq) + OH-(aq)

Kb = Kw/Ka = (1.0 x 10-14)/(8 x 10-5) = [HC6H7O6][OH-]/[ C6H7O6-]

1.25 x 10-10 = [OH-]2/0.0231

[OH-]^2 = 2.9 x 10-12,

[OH-]= 1.7 x 10-6
pOH = 5.77,
pH = 14- 5.77 = 8.23

(c) hypoiodous acid (HIO), Ka = 2.3e-11;
similarly we can calculate;

[IO-] = MbaseMacid/(Mbase+Macid)
[IO-] = (0.043)(0.050)/(0.043+0.050) = 0.0231 M

IO-(aq) + H2O(l) <====> HIO (aq) + OH-(aq)

Kb = Kw/Ka = (1.0 x 10-14)/(2.3 x 10-11) = [HIO][OH-]/[IO-]

4.35 x 10-4 = [OH-]^2/0.0231

[OH-]^2 = 0.1 x 10-4,

[OH-]= 3.16 x 10-3
pOH = 2.5,
pH = 14- 2.5 = 11.5