Question

TITRATION OF TUMS TABLETS

Using the data in table 5.2 and the mass of the tablets, calculate the mass and percent mass of calcium carbonate in each tablet. Clearly show your calculations below. Perform the calculations for each sample individually.

Mass of Tablets:

Sample 1: 1.305g

Sample 2: 1.312g

Sample 3: 1.314g

Concentration of HCL Supplied: 0.4524M

Table 5.2

	Trail 1	Trial 2	Trail 3
Final Burette Reading	18.22	33.88ml	16.18ml
Initial Burette Reading	0.79	18.22ml	0.38ml
Total Volume Added	17.43ml	15.66ml	15.80ml

Note that 1.082M NaOH is the liquid in the burret and that 25 ml HCl of 0.4524M and 25 mL of water is the liquid that the tums tablets dissolved in

Answer 1

Estimation of amount of CaCO3 in tablet

Reaction,

CaCO3 + 2HCl ---> CaCl2 + H2O

Trial 1,

moles of HCl added = molarity x volume

                                = 0.4524 M x 0.025 L = 0.01131 mol

moles of NaOH reacting = 1.082 M x 0.01743 L = 0.01886 mol

[pl. note moles of NaOH can never exceed moles of HCl added in this experiment, therefore, the molarity of NaOH is incorrect here. We would do a representative calculation with 0.1082 M NaOH]

moles of NaOH reacting = 0.1082 M x 0.01743 L = 0.001886 mol

moles of HCl reacted with CaCO3 = 0.01131 - 0.001886 = 0.009424 mol

moles of CaCO3 present = 0.009424 mol/2 = 0.004712 mol

mass of CaCO3 present = 0.004712 mol x 100.086 g/mol = 0.472 g

mass of tablet = 1.305 g

percent CaCO3 in tablet = 0.472 g x 100/1.305 g = 36.17%

Similarly other trial runs can be done