Question

Extra Credit: The Titration of Na2CO3 with HCl (25 pts) (You must show work to earn credit!) 1. A 0.1983 g sample of Na2CO3 was dissolved in 100.00 mL H20. That solution was titrated with 0.1531 M HCl as titrant. The pKai and pKaz of carbonic acid (the conjugate acid of the carbonate ion) are 6.351 and 10.329, respectively. (a) (2.5 pts) Write the equilibrium expressions for Kai and Kaz, using H2A, HA, and A2-for the chemical species. (b) (2.5 pts) What are the values for pKbi and pKuz? (C) (5 pts) What is the initial pH of the sodium carbonate solution? (e) (5 pts) What is the first equivalence point (in mL HCI)? What is the pH at 1st eq.? (f) (7.5 pts) What is the second equivalence point (in mL HCI)? What is the pH at 2nd eq.? (g) (2.5 pts) What is the pH when 18.33 mL of HCI have been added?

Answer 1

Multiple sub-parts, student have not specified questions as per guideline only first 4 sub-parts will be answered :

1.

(a) Equilibrium expression :

H₂CO₃ (aq) $\rightleftharpoons$ H⁺ (aq) + HCO₃^- (aq) ; Ka1 = [HCO₃^-] [ H⁺] / [ H₂CO₃]

HCO₃^- (aq)  $\rightleftharpoons$ H⁺ (aq) + CO₃^2- (aq) ; Ka2 = [CO₃^2-] [ H⁺] / [ HCO₃^-]

(b)

we have , pKw =pKw + pKb

So, pKb1 = pKw - pKa1 = 14.00 - 6.351 = 7.649

  pKb2 = pKw - pKa2 = 14.00 - 10.329 = 3.671

(c) Initial pH of Na₂CO₃ solution :

0.1938 of Na₂CO₃ added to 100.0 ml of water

Molar mass of Na₂CO₃ = 106 g /mol ; moles of Na₂CO₃ = 0.00183 moles

Solution strength = 0.0183 M = [ CO₃^2-]

on addition of it to water we have following equilibrium :

CO₃^2- + H₂O  $\rightleftharpoons$  HCO₃^- + HO^- ;  Kb = [ HCO₃^-] [ ^-OH]/ [ CO₃^2-] = 2.133*10^-4

we have [ HCO₃^-] = [ ^-OH]

[ ^-OH]² / [ CO₃^2-] = 2.133*10^-4

[ CO₃^2-] =  0.0183 M

or  [ ^-OH] ⁼ $\sqrt{}$ (2.133*10^-4  * 0.0183 ) = 1.975 * 10^-3

pH = pkw -pOH = 14.00 - (-log[ ^-OH] ) = 14.00 + log[ ^-OH] = 11.29

(e)

At First equivalence point all of CO₃^2- will be converted to  HCO₃^-.

mole of CO₃^2- = moles of HCl

so, moles of HCl required : 0.00183 M

Molarity of HCl = 0.1531 M = 0.1531 mol/ L

So, volume of HCl solution required = 12 ml.

pH at first equivalence point :

Equilibrium at this point,

H⁺ (aq) + CO₃^2- (aq)    $\rightleftharpoons$ HCO₃^- (aq) pka2 = 10.329 ; ka2 = 4.69*10^-11

At this point we have another equilibrium;

HCO₃^- + H₂O  $\rightleftharpoons$  H₂CO₃+ HO^- pka1 = 6.351 ; ka1 = 4.46*10^-7

,From simplified equation for weak diprotic acid [ H⁺ ] = $\sqrt{}$(ka1 *ka2) = 4.57*10^-9

^or pH = 1/2 (pka1 +pka2 ) = -log[ H⁺ ] = 8.34