Question

Assume that 22.00 mL of NaOH (0.100 M) was consumed to titrate a 10.00 mL H2CO3 solution (fully converted to Na2CO3) pKa1= 6.351 and pKa2= 10.329.

(a) what is the concentration of the initial H2CO3 solution?

(b) what is the pH of the H2CO3 solution before titration?

(c) which is the principal (dominating) species when 5.00 mL of NaOH has been added?

(d) what is the pH of the solution when 15.00 mL of NaOH has been added?

(e) what are the two most abundant species at pH=10.329?

Answer 1

In the first part , simple law of chemical equivalence is applied. In the second part we used one valid approximation . Because H ion comes only from the ist dissociation . In the 4th part there is formation of buffer solution . And we calculate ph by applying handerson equation.
Molarity of Naon = 0.0 M pkal pka 2 of of H₂CO3 = H₂CO3 = 6.351 10.329 Volume of Naon consumed = 22.00 me of H₂CO3 = lo nome Volume Law Apply of Chemical equivalence (Naoh) (14, CO₂) Nov. n, my, a = No V₂ n M₂ Vz 9x Mx10= I x 0.1892 Molanity of : 22 = 0.11 M Haraz Note! n factor for is H₂ coz ht for g. Because H₂CO3 I molecule of H₂CO3 can give 9 Molarity - H₂CO3 of all M pH of M2 (03 befuse titration (5) H₂CO3 = ht - Incoj itt com Heog - in comparison Because a to bkal pka2 is . There foll very we small can make one

o valid all Fxom approximation here H+ came from ist and dissociation is we can consider dissociation. He negligible. that came Thus, H₂CO3 H+ + HCO, Initially At equilibrium ((1-x) C2 Keq= Ea ca 6(1-2) Because keq is very small (2= seker [147) seker log (H')= -log (( Kerjar - 13 [ log c + dog Kee] pH = 1/3 loge - 3 log ker - - logo + p pkk!

pha [ pkai - lage] = 3 [6.351 - log (0.11)] (pH = 3.654] For full conversion (0) to the reonued Na2CO3 is 22.0 me Volume of Naon of Naon vol. for to n to vol Nanco, conversion Required Namco to me is 5 me Naon is . are Therefore when dominating species in Nanco, in K2CO3 (d) Volume of Naon added = 15.oo me the of 11.00 me Conversion Naon of is Haco, consumed in Nancoz. Remaining 4.50 me is used for Nanco, Nazco, No + H2O Narcos Naon = No. of moles of Naonco3 = 0.l1x 10 11 mol No. of moles uxou-o.u mal of For Naon this consumed step No. of moles of Nancoz deft: Hoon 0.7 mol -

It will e its PR equetion taim forma can be And a buffer calculated solution. Handerson by pH = pkaz & lag [co, [mece, Ipu - 10.086 ce) abundant species Two most will be in Nallcaz (in Naz (og