A 0.5250 sample containing only NaHCO_3 (MW = 84.01) and Na_2CO_3 (MW = 105.99)

Question

A 0.5250 sample containing only NaHCO_3 (MW = 84.01) and Na_2CO_3 (MW = 105.99) was dissolved in 30.0 mL of water. The solution was titrated with 0.2010 M HCI and the first equivalence point in the reaction was reached when 14.20 mL of acid had been added. Ka_1(H_2CO_3) = 4.3 times 10^-7 and Ka_2(H_2CO_3) = 4.8 times 10^-11. Determine the weight percent of Na_2CO_3 in the original sample. Determine the total volume of acid needed to reach the second equivalence point. Calculate the pH of the solution at the beginning of the titration.

Explanation / Answer

Solution:

a) At the beginning of the titration you have a mixture of CO32- + HCO3-.

At the first titration equivalence point the volume of the acid added is 14.2 mL. The reaction occurring at the equivalence point of titration is as follows:
CO32- + H+ ----> HCO3-.

Percent Na2CO3 can be calculated at the equivalence point as follows:

g of Na2CO3 = mL x M x milli-equivalent weight = 14.2 x 0.2010 x 0.106 = 0.3025g Na2CO3 = 0.0028 mol

Therefore, % Na2CO3 = (g Na2CO3/mass sample)x100 = (0.3025/0.5250)x100 = 57.6%

The next part of the titration titrates what is left i.e all of the CO32- (which is now HCO3-) + HCO3- there from the initial. The volume needed to titrate the carbonate (14.2 mL x 2)= 28.4 mL (It take 14.2 mL to titrate the carbonate half way so it takes 28.4 mL to titrate it all the way).

g of HCO3- = 14.2 x 0.2010 x 0.084 = 0.2397g = 0.0028 mol

c) Concentration of Na2CO3 =[CO32-]= 0.0028 mol/0.03 L = 0.093 M

Concentration of NaHCO3 =[ HCO3-]= 0.0028 mol/0.03 L = 0.093 M

The solution is made of NaHCO3 and Na2CO3 then the following equation will happen in solution

HCO3- (aq) + OH- (aq) CO32- (aq) + H2O(l) ---------(1)

CO32- (aq) + H+ (aq) HCO3- (aq)

Now, Using Henderson-Hasselbalch equation for (1):

pH = pKa2 + log ([CO32-]/[ HCO3-])

pH = pKa2                    (since, [CO32-]=[ HCO3-]=0.093 M)

pH = pKa2 = -logKa2

pH = - log(4.8x10-11)= 11 – 0.68 = 10.31

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