Question

The following question pertains to a back titration used to determine the calcium in a Tum's tablet

Calculate the amount of KHP (Mw= 204.22 g/mol) necessary to reach the endpoint by titration with 25 mL of approx 0.1 M NaOH

The expected amount of CaCO3 within a Tum's antacid tablet is 500 mg. Calculate the amount of 0.0996 M HCl required to reach a back titration (using 0.1064 M NaOH). Aim to end the titration after 25 mL addition of NaOH.

Answer 1

Reaction

KHP + NaOH ---> NaKHP + H2O

Concentration of NaOH = 0.1 M

Volume of NaOH = 25 mL

No. of moles of NaOH = 0.1 M x 25/1000 mL = 0.0025 mol

Molar mass of Potassium hydrogen phthalate KHP = 204.22 g/mol

Mass of 0.0025 mol KHP = 0.0025 mol x 204.22 g/mol = 0.5106 g

Amount of KHP required = 0.5106 g

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Amount of CaCO3 = 500 mg = 0.500 g

Molar mass of CaCO3 = 100.09 g/mol

No. of moles of CaCO3 = 0.500 g/100.09 g/mol = 0.005 mol

Reaction

CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O

Back titration

HCl + NaOH ---> NaCl + H2O

From the balanced equation each mole of CaCO3 reacts with 2 moles of HCl in the reaction

Once you add CaCO3 to X mL of 0.0996 M (x mol) HCl, it will use 2 x 0.005 = 0.01 mol of HCl.

No. of moles of NaOH in 25 mL, 0.1064 M = 0.1064 M x 25/1000 mL = 0.0027 mol

So no. of moles of HCl required in the solution for the reaction with CaCO3 is 0.01 mol and 25 mL NaOH is 0.0027 mol

Total moles of HCl need to take = 0.01 + 0.0027 = 0.0127 mol

Concentration of HCl = 0.0996 M

Volume of HCl required to make 0.0127 mol sol = 0.0127 mol / 0.0996 mol/L = 0.12751 L = 127.51 mL

So the volume of HCl required to take for titration is more than 127.5 mL