Chapter 4, Problem 122IL

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# In some laboratory analyses, the preferred technique is to dissolve a sample in an excess of acid or base and then “back-titrate” the unreacted acid or base with a standard base or acid. To assess the purity of a sample of (NH4)2SO4 you dissolve a 0.475-g sample of impure (NH4)2SO4 in aqueous KOH.(NH4)2SO4(aq) + 2 KOH(aq) → 2 NH3(aq) + K2SO4(aq) + 2 H2O(ℓ)The NH3 liberated in the reaction is distilled from the solution into a flask containing 50.0 mL of 0.100 M HCl. The ammonia reacts with the acid to produce NH4Cl, but not all of the HCl is used in this reaction. The amount of excess acid is determined by titrating the solution with standardized NaOH. This titration consumes 11.1 mL of 0.121 M NaOH. What is the weight percent of (NH4)2SO4 in the 0.475-g sample?

Interpretation Introduction

Interpretation:

The weight percent of (NH4)2SO4 in the given sample has to be determined.

Concept introduction:

• The relation between the number of moles and mass of the substance is ,

Numberofmole=Mass in gramMolarmass

Mass in gramofthe substance=Numberofmole×Molarmass

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Amountof substance=Concentrationofsubstance×volumeofthesubstance
• Weight percent or mass percent of elements of a compound is the ratio of weight of element to the weight of whole compound and multiplied with hundred.
Explanation

The amount of hydrochloric acid in 50.0â€‰mLâ€‰â€‰ofâ€‰â€‰0.100â€‰Mâ€‰HCl can be determined as follows,

â€‚Â Amountâ€‰ofÂ HCl =â€‰Concentrationâ€‰HClâ€‰Ã—volumeâ€‰HCl =0.100â€‰â€‰molLâ€‰Ã—0.05â€‰L =â€‰â€‰0.005â€‰â€‰molesÂ â€‰ofâ€‰â€‰HCl

Therefore the initial amount of HCl is 0.005â€‰â€‰mol

Amount of NaOH used in reaction with HCl is 0.0013â€‰mol

It is because 1000â€‰â€‰mL contains 0.121â€‰moles, then

11.1â€‰mLâ€‰(usedâ€‰inâ€‰reactionâ€‰withâ€‰HCl)â€‰â€‰ contains â€‰0.1211000â€‰Ã—11.1â€‰=â€‰0.0013â€‰moles

The balanced equation for the reaction of HClâ€‰â€‰andâ€‰â€‰NaOH is,

â€‚Â HCl+NaOHâ€‰â†’NaCl+H2O

The stoichiometric ratio between the reactant NaOH and HCl is 1:1

Therefore 0.0013 moles of NaOH reacts with 0.0013 moles of HCl.

So the final amount of HCl is 0.0013 moles.

Amount of HCl reacted with NH3 is,

Initialâ€‰â€‰amountâ€‰â€‰ofâ€‰HClâ€‰âˆ’â€‰â€‰Finalâ€‰amountâ€‰â€‰ofâ€‰â€‰HCl = 0.005â€‰â€‰-â€‰â€‰0.0013 =â€‰â€‰0.0037â€‰â€‰moles

Here to assess the purity of a sample of (NH4)2SO40

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