Solid carbonates and bicarbonates will react with strong acids to produce salt,

Question

Solid carbonates and bicarbonates will react with strong acids to produce salt, carbon dichloride gas, and water. Adding a solid carbonate or bicarbonate to a strong acid will produce bobbies, and once the acid is neutralized the bubbling will stop. Slowly adding a solid carbonate or bicarbonate to a known volume and concentration of a strong acid is similar to a titration, that is you can use the known moles of the strong acid to determine the moles of the carbonate or bicarbonate. You are given 1.546 grams of a white powder and told that it is a mixture of potassium carbonate and sodium carbonate. You are also asked to determine the percent composition [percent sodium and potassium carbonate) by mass of the sample. To find the percent composition you decide to slowly add the powder to 10.00 mL of 1.15 M nitric acid until the bubbling stops. After adding the powder to the acid you reweigh your sample and its mass is 0.802 g. What is the percent composition by mass of the powder?

Explanation / Answer

Ans. Let the mass of Na2CO3 = A gram , and mass of K2CO3 = B gram.

So, total mass of the mixed salt-

            A g + B g = 1.546 g

            Or, A + B = 1.546      - equation 1

Now,

Moles of Na2CO3 in sample = mass / molar mass = A g/ (105.98 g/mol) = (A / 105.98) mol

Moles of K2CO3 in sample = mass / molar mass = B g/ (138.20 g/mol) = (B / 138.20) mol

#2: Acid treatment:

Moles of HNO3 = Molarity x Volume (in L)

                        = 1.15 M x 0.010 L                                          ; [1 L = 1000 mL]

                        = (1.15 mol/ L) x 0.010 L                                 ; [1 M = 1 mol/ L]

= 0.0115 mol

Stoichiometry: 1 mol of each reactant release 1 mol CO2 and H2O upon acid treatment.

Mass of CO2 produced = Initial mass of mixture – Mass remaining after acid treatment

                                    = 1.546 g – 0.802 g = 0.744 g

During acid treatment CO2 is lost result loss in the mass of initial sample. It’s also assumed that mass loss is in form of CO2 and H2O only.

Let the moles of CO2 produced = X mol

So, moles of H2O produced = X mol, too.

Total mass lost = (X mol x molar mass of H2O) + (X mol x molar mass of CO2)

Or, 0.744 g = 18.01 X g + 44.0 X g

Or, X = 0.011998

Hence, moles of CO2 lost = 0.011998 mol

Assuming the reaction to be complete CO2 removal (until bubble stops), the moles of CO2 produced is equal to-

175.72888863086598935655539429124

            Moles of CO2 produced = (1) x moles of Na2CO3 + (1) x moles of Na2CO3

            Or, 0.011998 mol = (A / 105.98) mol + (B / 138.20) mol

            Or, 0.011998 x 14646.436 = 138.20 A + 105.98 B

            Or, 138.20 A + 105.98 B = 175.23               - equation 2

Comparing (equation 1 x 105.98) – equation 2

            105.98A + 105.98B = 163.845

       (-) 138.20 A + 105.98 B = 175.23

            Or, -32.22 A = -11.385

            Or, A = 11.385/ 32.22 = 0.35335

Thus, mass of Na2CO3 = A g = 0.35335 g

Mass of K2CO3 = BA g = 1.546 - 0.35335 g = 1.19265 g

% of Na2CO3 = (Mass of Na2CO3 / total mass of mixture) x 100

                        = (0.35335 g / 1.546 g) x 100

                        = 25.856 %

% of K2CO3 = (Mass of K2CO3 / total mass of mixture) x 100

                        = (1.19265 g / 1.546 g) x 100

                        = 74.144 %

Note: This is NOT a good quality question as it does not specify the way the product was measured (dry product or mass in the beaker itself). However, to get result, it’s the weight after reaction is that of dry mass. Please ask your curiculum to specify it.

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