The active component in one type of calcium dietary supplement is calcium carbonate. A 1.2450 g tablet of the supplement is added to 65.00 mL of 0.4984 M HCl(aq) and allowed to react. After completion of the reaction, the excess HCl(aq) requires 37.15 mL of 0.2065 M NaOH(aq) for its titration to the equivalence point. What is the calcium content of the tablet, expressed in milligrams of Ca?

The calcium content of the tablet is 486 mg.

The reaction between the HCl and the supplement occurs as follows;

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

Total amount of HCl reacted = 65/100 L × 0.4984 M = 0.032 moles

From the reaction;

HCl(aq) + NaOH(aq) ---> NaCl (aq) + H2O(l)

Number of moles of NaOH = 37.15/1000 × 0.2065 M = 0.0077 moles

Since the reaction is 1:1, 0.0077 M of excess HCl is present

Amount of HCl reacted = 0.032 moles - 0.0077 moles = 0.0243 moles

1 mole of calcium carbonate reacts with 2 moles of HCl

x moles of calcium carbonate reacts with 0.0243 moles of HCl

x = 1 mole ×  0.0243 moles/2 moles

x = 0.01215 moles

Mass of CaCO3 reacted =  0.01215 moles × 100 g/mol = 1.215 g

If 100 g of CaCO3 contains 40 g of Ca

1.1215 g of CaCO3 contains  1.215 g × 40 g/100 g = 0.486 g or 486 mg

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Explanation:

This is an example of back titration.

You add a known excess of HCl to the CaCO₃. Some of it reacts with the CaCO₃.

You titrate the left-over HCl.

From the difference you can calculate the amount of CaCO₃.

         100.09

         CaCO₃ +  2HCl ⟶ CaCl₂ + H₂CO₃

m/g:   1.2450

V/mL:                 65.00

c/mol·L⁻¹:          0.4984

                  NaOH + HCl ⟶ NaCl + H₂O

V/mL:           37.15

c/mol·L⁻¹:   0.2065

1. Total moles of HCl

2. Excess moles of HCl

3. Moles of HCl that reacted

n = (32.40 - 7.671) mmol HCl = 24.72 mmol HCl

4. Moles of CaCO₃

5. Mass of Ca

thats how its supposed to be

it takes on an electronical energy