Question

Titration of a diprotic acid with a strong base You have a 10.0 mL solution containing 0.5 M carbonic acid. Carbonic acid is diprotic, with pKa1 = 6.35 and pKa2 = 10.33. You titrate this solution using 1.00 M NaOH .

(a) Calculate the pH of the solution before adding any NaOH.

(b) Calculate the amount of NaOH needed to reach the first midpoint. What is the pH?

(c) Calculate the amount of NaOH needed to reach the first equivalence point. What is the pH?

(d) Calculate the amount of NaOH needed to reach the second midpoint. What is the pH?

(e) Calculate the amount of NaOH needed to reach the second equivalence point. What is the pH?

(f) Calculate the pH of the solution after adding 15 mL of NaOH.

(g) Use your above calculations to sketch the pH for the solution during the entire titration.

Answer 1

Ž HCO3 + H+ Ango - H₂CO3 prag = 6.35. - KO . Ka = 4.407 107 HCO3 co²- + Ht i n - pka = 10.33 Kl = 4.67 x 10" Let H₂ W3 = H₂A (for simplicity so too ml of 0.5M H₂A. (a) H₂A H+ + HA pka, 6.35 0.5 do o Ka = 4.447 0107 0.5-x Z Ka = 4.47x10 t = x2 -0.5-x 2 0.5 > > X. - as x² = 0.5 x 4,47 xlo7 x = 4.7x10-4 M by and dissociation it will be negligible as pka, >> pka₂. Ignore (H+] MA ou pH = -log(Ht] - - log (4.9 X604) pH = 3.327

(6 At mild of pt. equivalance point. [H₂A) = [Na] pH = pka, = 6.35 u volume of IM NaOH is needed M,V, = M2 V2 1 = 0.5 x 10 ml 2 V = 2.5 ml. (c) At ist equi valance point. HA is major species both as acid and base - Camphoteric species) pH = (pka, + pka2)/2 = 6,35+10:33 pH = No ret 8.34 volume of lu NaOH MV, a M2 V2 is needed

1x V ml = V 0.5x 10 = 5 ml (d) At point mid of and equivalance - THAT = [42 ] pH = pka, pH = 10.33 Let Ve volume of IM NaOH is needed Lo Nc = 6 + 10.5*100 Ve = 7.5 mt (es I At 2nd equivalance point only A²- present, (weak base) A² + H₂O HA + OH Kis Kw = 10014 o kar 4.6781011 Kb = 2.148104 volume of NaOH needed = 22 Volume of M₂ AxMest o molarity of NaOH Vd = 2x 10 x 0.5 ml Here Tot al volume dy mmol of TA2 Vd & to me.. solution = 20 nl. A² = 0x10 = 5 mmol. ] = 5 mmol = 0,25M 20 me

Fage no 42- + 0.25 H₂O - & HA o OH one 0.25-x k = 2.14x10 4 - x² O 0.25-2 xca 0.25 x= 0.25 x 2.14x10-4 x = 0.73x102" N = 7.3 x 10-3 u. [or] = 7.3 x 103 u polt = -log (OH-] = -log (7. 3x10 3). poH = 2.136 pH = 14.2.136 pH = 11.26