Question

You weighed a solid sample of an unknown mixture of carbonate and bicarbonate to be 5.3431 grams. You properly dissolved the entire unknown solid sample in 250.0 mL of boiled and cooled distilled water. You then performed two trials of 25.00 mL aliquot of the unknown solution titrated with 0.1389 M standardized HCl using a pH probe yielding: trial 1:{volume HCl (mL), pH} = {{1.090, 12.15}, {9.190, 12.13}, {17.290, 12.33}, {25.590, 12.35}, {33.800, 12.13}, {41.830, 12.30}, {49.990, 11.34}, {58.280, 7.10}, {66.500, 3.21}, {74.580, 2.57}, {82.760, 2.50}, {91.030, 2.53}, {99.070, 2.45}, {107.360, 2.52}}, and trial 2: {volume HCl (mL), pH} = {{0.900, 10.37}, {9.220, 10.49}, {17.320, 10.34}, {25.530, 10.35}, {33.720, 10.39}, {41.980, 10.22}, {49.990, 9.46}, {58.240, 5.32}, {66.460, 1.81}, {74.680, 1.22}, {82.880, 1.19}, {90.990, 1.18}, {99.210, 1.19}, {107.320, 1.16}}. Then you got really crazy and decided to do two more trials by adding 50.00mL of 0.1301 M standardized NaOH and then excess barium chloride to precipitate all remaining carbonate. The excess hydroxide was titrated for each trial with the above standardized HCl solution using a conductivity probe yielding: trial 1: {volume HCl (mL), Conductivity (µS/mL)} = {{0.00, 289814}, {2.41, 283202}, {4.80, 271022}, {7.14, 255543}, {9.51, 236964}, {12.00, 214753}, {14.36, 191397}, {16.80, 165119}, {19.18, 137574}, {21.25, 112148}, {23.74, 79875}, {26.08, 47950}, {29.29, 5151}, {31.65, 5545}, {34.15, 4567}, {36.37, 2537}, {39.18, -1542}, {41.28, -5658}, {43.89, -12009}, {46.47, -19593}, {48.56, -26663}, {50.92, -35621}, {53.77, -47776}, {55.98, -58194}}, and trial 2: {volume HCl (mL), Conductivity (µS/mL)} = {{0.00, 269987}, {2.40, 264037}, {4.74, 253103}, {7.10, 238729}, {9.49, 221376}, {11.97, 200831}, {14.26, 179800}, {16.84, 153956}, {19.11, 129476}, {21.33, 104058}, {24.00, 71650}, {26.36, 41437}, {29.73, 8778}, {32.19, 21978}, {34.43, 33118}, {36.76, 43848}, {39.29, 54540}, {41.43, 62828}, {43.77, 71118}, {46.62, 80163}, {48.12, 84472}, {50.81, 91424}, {53.45, 97325}, {56.20, 102500}}.

Calculate the average percent weights of the bicarbonate and carbonate in the original sample.

Answer 1

Calculation

Draw graph between Volume of HCl vs Ph and find the end point of titration from mid point of graph

Experiment 1

Trial 1

End point, volume of HCl = 58.280 ml

Normality of HCl      = 0.1389 N

Volume of carbonate and bicarbonate mixture = 25 ml

Normality of carbonate and bicarbonate mixture = (58.280×0.13890)/25

                                                                              = 0.3238 N

Trial 2

End point, volume of HCl = 58.240 ml

Normality of HCl      = 0.1389 N

Volume of carbonate and bicarbonate mixture = 25 ml

Normality of carbonate and bicarbonate mixture = (58.240×0.13890)/25

                                                                              = 0.3235 N

Result

Mean Volume of HCl = 58.26 ml

Normality of mixture = 0.3236 N

Experiment 2 (Excess NaOH Vs HCl)

Trial 1

Volume of HCl = 29.29 ml

Trial 2

Volume of HCl = 29.73 ml

Mean volume = 29.51 ml

Volume of HCl that react with excess NaOH = 29.51 ml

Calculation

First 25 ml of carbonate and bicarbonate mixture require 58.26 ml of HCl for reaction

Let total mixture contain x mol of carbonate and y mole of bicarbonate

Total alkalinity = [ HCO₃^-] + 2 [CO₃^2-] which is equal to the mol of HCl required to reach the end point

Mol of HCl required to reach the end point = 0.13890 × 58.26

                                                                      = 8.09 mol

[ HCO₃^-] + 2 [CO₃^2-]                                     = 8.09 ml

Mol of HCl in experiment 2                          = 0.13890 × 29.29             

                                                                       = 4.06 mol

Mol of NaOH = 50×0.1301= 6.505

Mol of NaOH reacts with HCO₃^-                       = 6.505-4.06

                                                                       = 2.445 mol

[ HCO₃^-] + 2 [CO₃^2-]                                     = 8.09 ml

2 [CO₃^2-]                                     = 8.09 ml - [ HCO₃^-]

                                                    = 8.09-2.445

                                                    = 5.645

                                    [CO₃^2-]   = 5.645/2

                                                   = 2.8825

Concentration of carbonate = 2.8825 mol

Concentration of bicarbonate = 2.445 mol