A sample of antacid tablet was titrated to determine its CaCO_3 content First th

Question

A sample of antacid tablet was titrated to determine its CaCO_3 content First the tablet was added with 5 mL of 3 0 M HCl and diluted with 50 mL of water Using phenolphthalein as the indicator, the solution was titrated to a color change of colorless to pink. It took 7.8 mL of 0.30 M NaOH to neutralize the excess HCl Calculate the mass of CaCO_2 in the tablet The balanced chemical equations for the two reactions in this titration are given below 2 HCI + CaCO_3 rightarrow CaCl_2 + H_2O + CO_2 HCI + NaOH rightarrow NaCl + H_2O Calculate the mass % of acetic add (CH_3COOH) in vinegar if it takes 16 7 mL of 0 15 M NaOH to neutralize a 3.05 g sample of the vinegar CH_3COOH + NaOH rightarrow CH_3COONa + H_2O Mass % acetic add = (mass acetic acid + mass vinegar) times 100

Explanation / Answer

Q1)

no of mole of HCl added = M*V = 5*3 = 15 mmole

no of mole of NaOH added = 7.8*0.3 =   2.34 mmole

no of mole of HCl excess = no of mole of NaOH added = 2.34 mmole

no of mole of HCl reacted = 15 - 2.34 = 12.66 mmole

so that

CaCo3 + 2HCl ----> CaCl2 + H2O + CO2

1 mole CaCo3 = 2 mole HCl

no of mole of CaCo3 present = 12.66/2 = 6.33 mmole

mass of CaCo3 = n*mwt = 6.33*10^-3*100 = 0.633 g

Q2)

   CH3COOH + NaOH ------> CH3COONa + H2O

1 mol CH3COOH = 1 mol NaOH

no of mole of NaOH reacted = 16.7*0.15 = 2.505 mmole

no of mole of CH3COOH reacted = 2.505 mmole

mass of CH3COOH reacted = 2.505*10^-3*60 = 0.15 g

mass% of aceticacid = mass of CH3COOH / sample mass * 100

                  = 0.15/3.05*100

                  = 4.92 %

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