The amount of CaCO_3 per antacid tablet was determined by dissolving a sample of

Question

The amount of CaCO_3 per antacid tablet was determined by dissolving a sample of antacid (0.145 g) in 10.00 mL HCl 0.500 M followed by back titration with 15.0 mL of standardized solution of NaOH with a concentration of 0.250 M. The original whole antacid tablet weighed 1.400 grams. a) Write down the balanced equation between CaCO_3 and HCI: b) Write down the balanced equation between NaOH and HCI: c) Explain why the back titration technique was used to determine the amount of CaCO_3 in the antacid tablet rather than performing a direct titration with HCI. d) Fill in the missing data to complete the Data Table shown below. Show your calculations in the table where necessary.

Explanation / Answer

Ans. A. CaCO3(s) + 2HCl(aq) -----------> CaCl2(aq) + CO2(g) + H2O(l)

#B. NaOH(aq) + HCl(aq) -----------> NaCl(aq) + H2O(l)

#C. The antacid is dissolved in a solution with excess of HCl. Some of HCl is neutralized as (reaction #A) while the remaining is neutralized following reaction (#B).

Since reaction #B is done against standard NaOH, the amount of HCl neutralized by it is calculated using C1V1. The amount of HCl neutralized by CaCO3 is calculated by subtracting moles of HCl neutralized from total moles of HCl used to dissolve the tablet following correct stoichiometry.

# D1. 1.400 g

#D2. 0.145 g

#D3. 10.0 mL = 0.010 L                                                                    ; [10 mL = 0.010 L]

#D4. Moles of HCl = Molarity x Volume of HCl in liters

                                    = 0.500 M x 0.010 L                                    

                                  = (0.500 mol/ L) x 0.010 L                           ; [1 M = 1 mol/L]

                                    = 0.005 mol

#D5. 15.50 mL = 0.0155 L

#D6. Moles of NaOH= Molarity x Volume of NaOH in liters

                                    = 0.250 M x 0.0155 L                                   ; [10 mL = 0.010 L]

                                  = 0.003875mol

#D7. See reaction #B. 1 mol NaOH neutralizes 1 mol HCl.

Thus, moles of HCl neutralized by NaOH = Moles of NaOH used to reach endpoint

                                                = 0.003875mol

#D8. Moles of HCl neutralized by tablet =

                        (Total moles of HCl, #D4) – (Moles of HCl neutralized by NaOH, #D7)

                        = 0.005 mol – 0.003875 mol

                        = 0.001125 mol

#D9. See reaction #A.

2 mol HCl neutralize 1 mol CaCO3.

So,

Moles of CaCO3 = (1/2) x Moles of HCl neutralized by tablet

                                    = (1/2) x 0.001125 mol

                                    = 0.0005625 mol

#D10. Mass of CaCO3 in the sample of antacid tablet = Moles x Molar mass

                                    = 0.0005625 mol x 100.09 g/mol

                                    = 0.056300625 g

#D11. Mass of CaCO3 in the whole antacid tablet = CaCo3 mass in the sample

                                    = 0.056300625 g

#D12. % CaCO3 = (Mass of CaCO3/ mass of tablet) x 100

                                    = (0.056300625 g / 0.145 g) x 100

                                    = 38.83 %

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