Question

The active component in one type of calcium dietary supplement is calcium carbonate. A 1.2450-g tablet of the supplement is added to 50.00 mL of 0.5000 M HCl (aq) and allowed to react. After completion of the reaction, the excess HCl (aq) requires 40.65 mL of 0.2535 M NaOH (aq) for its titration to the equivalence point. What is the calcium content of the tablet, expressed in milligrams of Ca^2+? m_ca^2+ = mg

Answer 1

Initial moles of HCl in 50 mL of 0.5 M HCl = 0.05 L x 0.5 M

= 0.025 moles

Now,

Excess moles of HCl is neutralized with NaOH. So, moles of NaOH = moles of HCl

Moles of NaOH in 40.65 mL of 0.2535 M NaOH = 0.04065 L x 0.2535 M

= 0.010 moles

So, excess moles of HCl = 0.010 moles

Now, the moles of HCl reacted with Calcium = 0.025 moles - 0.010 moles = 0.015 moles

The balanced molecular equation for the reaction of CaCO3 and HCl is as follows:

CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

Or

CaCO3(s) + 2H⁺(aq) + 2Cl^-(aq) → Ca²⁺(aq) + 2Cl^-(aq) + H₂O(l) + CO₂(g)

Or

CaCO3(s) + 2H+(aq) → Ca²⁺(aq) + H2O(l) + CO2(g)

Here,

2 moles of H+ produces 1 mole of Ca²⁺

$\Rightarrow$ 1 moles of H+ produces 1/2 mole of Ca²⁺

$\Rightarrow$ 0.015 moles of H+ produces (0.015/2) mole of Ca²⁺

$\Rightarrow$ 0.015 moles of H+ produces 0.0075 mole of Ca²⁺

So, there are 0.0075 mole of Ca²⁺.