Question

a) The antacid component of Tumsr is calcium carbonate. Assume Tumsr is 40.0 percent CaCO3 by mass. If we have 400. mg of Tumsr how many ml of 0.100 M HCl can we neutralize? Express your answer in mL.

b) For every 5.00 mL of milk of magnesia there are 400. mg of magnesium hydroxide. How many mL of milk of magnesia do we need to neutralize 40.0 mL of 0.500 M HCl? Express your answer in mL.

c) Suppose 13.00 mL of 0.100 M barium hydroxide is required to neutralize 17.00 ml of nitric acid with an unknown concentration. What is the concentration of the nitric acid. Express your answer in mol/L.

Answer 1

a) Tumsr has 40.0% CaCO₃ by mass.

Therefore, mass of CaCO₃ in 400 mg Tumsr = 40.0/100*(400 mg) = 160 mg.

Write down the balanced equation of reaction between CaCO₃ and HCl.

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

The molar ratio of reaction between CaCO₃ and HCl is 1:2.

Molar mass of CaCO₃ = 100.1 g/mol.

Therefore, moles of CaCO₃ in 400 mg Tumsr = (160 mg)*(1 g/1000 mg)*(1 mole/100.1 g) = 1.5984*10^-3 mole (I have kept a few guard digits extra).

Therefore, moles of HCl neutralized as per the stoichiometric equation = (1.5984*10^-3 mole CaCO₃)*(2 mole HCl/1 mole CaCO₃) = 3.1968*10^-3 mole.

Therefore volume of HCl neutralized = mole(s) HCl neutralized/molarity of HCl = (3.1968*10^-3 mole/0.100 mole/L) = 0.031968 L = (0.031968 L)*(1000 mL/1 L) = 31.968 mL ≈ 32.0 mL.

Ans: The volume of HCl neutralized = 32.0 mL.

b) Write down the balanced neutralization reaction:

Mg(OH)₂ + 2 HCl -------> MgCl₂ + 2 H₂O

The molar ratio of reaction between HCl and Mg(OH)₂ is 2:1.

Moles HCl neutralized = (volume of HCl in L)*(concentration of HCl in moles/L) = (40.0 mL)*(1 L/1000 mL)*(0.500 mole/L) = 0.0200 mole.

Moles Mg(OH)₂ reacted = (0.0200 mole HCl)*(1 mole Mg(OH)₂/2 mole HCl) = 0.0100 mole.

Molar mass of Mg(OH)₂ = 58.32 g/mol.

Therefore, mass Mg(OH)₂ reacted = (0.0100 mole)*(58.32 g/mol) = 0.5832 g = (0.5832 g)*(1000 mg/1 g) = 583.2 mg.

Now, 5.00 mL milk of magnesia has 400 mg Mg(OH)₂.

Therefore, volume of milk of magnesia required = (583.2 mg)*(5.00 mL milk of magnesia/400 mg Mg(OH)₂) = 7.29 mL ≈ 7.3 mL

Ans: The volume of milk of magnesia required = 7.3 mL.

c) Write down the neutralization reaction:

Ba(OH)₂ + 2 HNO₃ -------> Ba(NO₃)₂ + 2 H₂O

The molar ratio of reaction between Ba(OH)₂ and HNO₃ = 1:2.

Moles Ba(OH)₂ added = (volume of Ba(OH)₂ in L)*(concentration of Ba(OH)₂ in moles/L) = (13.00 mL)*(1 L/1000 mL)*(0.100 mole/L) = 1.3*10^-3 mole.

Moles HNO₃ neutralized as per the balanced stoichiometric equation = (1.3*10^-3 mole Ba(OH)₂)*(2 mole HNO₃/1 mole Ba(OH)₂) = 2.6*10^-3 mole.

Concentration of nitric acid in moles/L = moles HNO₃ neutralized/volume of HNO₃ neutralized in L = (2.6*10^-3 mole)/[(17.00 mL)*(1 L/1000 mL)] = 2.6*10^-3*1000/(17.00*1) mole/L = 0.1529 mole/L ≈ 0.153 mole/L

Ans: The concentration of HNO₃ = 0.153 mole/L.