Question

1. Calculate the mass of KHC₈H₄O₄ (molar mass = 204.44g/mol) that reacts with 15mL of NaOH solution in part A.3. Express this mass of KHC₈H₄O₄ to the correct number of significant figures and record the calculation on the report sheet.

2. A 0.204g sample of CO₃^2- antacid is dissolved with 25.0mL of 0.0981M HCl. The hydrochloric acid that is not neutralized by the antacid is titrated to a bromophenol endpoint with 5.83mL OF 0.104M NaOH. Calculate the number of moles of base in the antacid.

Answer 1

Solution :-

Given data

Lets calculate the moles of the HCl

Moles of HCl = molarity * volume in liter

= 0.0981 mol perL * 0.025 L

= 0.0024525 mol HCl

Now lets calculate the moles of the NaOH

Moles of NaOH = molarity * volume in liter

= 0.104 mol perL* 0.00583 L

= 0.0006063 mol NaOH

Now moles of HCl reacted with the NaOH are same as moles of NaOH because mole ratio is 1 :1 for the HCl and NaOH

Therefore the moles of HCl reacted with the base in the antacid are calculated as

Moles of HCl reacted with antacid base = moles of HCl – moles of naOH

= 0.0024525 mol – 0.0006063 mol

= 0.0018462 mol

Now lets calculate the moles of base present in the antacid

Reaction equation is as follows

CO3^2-+ 2HCl----- > CO2 + H2O +2 Cl-

Therefore mole ratio of the CO3^2- to HCl is 1 :2

0.0018462 mol HCl * 1 mol CO3^2- / 2 mol HCl = 0.000923 mol CO3^2-

Therefore moles of base present in the antacid are 9.23*10^-4 mol