A solution containing 10.mmol of CO₃²⁻ and 5.0 mmol of HCO₃−is titrated with 1.1 M HCl.What total volume of HCl must be added to reach the second equivalence point? (1 mmol = 0.001 mol) A) 13.6 mL B) 9.1 mL C) 4.5 mL D) 22.7 mL E) 27.7 mL

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Quizplus · Accepted Answer

The titration equation for CO₃²⁻ and HCO₃− is: CO₃²⁻ + HCl → CO₃− + H2O + Cl− In the first equivalence point, all of the HCO₃− will be neutralized to CO₃−, so the number of moles of HCl required is: mol of HCO₃− = 5.0 mmol mol of HCl = 5.0 mmol Since the concentration of the HCl is 1.1 M, the volume of HCl required is: V = mol of HCl / [HCl] = 5.0 mmol / 1.1 mol/L = 4.5 mL At the second equivalence point, all of the CO₃²⁻ will be neutralized to CO₃−, so the total number of moles of HCl required is: mol of HCO₃²⁻ + mol of HCO₃− = 10.0 mmol + 5.0 mmol = 15.0 mmol The volume of 1.1 M HCl required to reach the second equivalence point is: V = mol of HCl / [HCl] = 15.0 mmol / 1.1 mol/L = 13.6 mL Therefore, the best answer is D.

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