Question

A 25.0 mL sample of sulfuric acid is completely neutralized by adding 32.8 mL of 0.116 mol/L ammonia solution. Ammonium sulfate, (NH4)2SO4, and water are formed. What is the concentration of the sulfuric acid?Select one:a. 0.608 mol/Lb. 0.152 mol/Lc. 0.304 mol/Ld. 0.08 mol/L

Answer 1

1) List the known and unknown quantities.

Reactant 1: Acid

Sulfuric acid: H2SO4

Volume: 25.0 mL.

Molarity: unknown (mol/L).

Reactant 2: Base.

Ammonia: NH3.

Volume: 32.8 mL

Molarity: 0.116 mol/L

2) Balance the chemical equation.

$H_2SO_4+NH_3\rightarrow(NH_4)_2SO_4$

List the elements in the reactants.

H: 5

S: 1

O: 4

N: 1

List the elements in the products.

H: 8

S: 1

O: 4

N: 2

3) Balance N.

$H_2SO_4+2NH_3\rightarrow(NH_4)_2SO_4$

List the elements in the reactants.

H: 8

S: 1

O: 4

N: 2

List the elements in the products.

H: 8

S: 1

O: 4

N: 2

4) Find moles of ammonia (NH3)

Molarity: 0.116 mol/L.

Volume: 32.8 mL = 0.0328 L

$mol\text{ }NH_3=0.116\frac{mol}{L}*0.0328\text{ }L=0.0038048\text{ }mol\text{ }NH_3$

5) Convert moles of NH3 to moles of H2SO4.

The molar ratio between NH3 and H2SO4 is 2 mol NH3: 1 mol H2SO4.

$mol\text{ }H_2SO_4=0.0038048\text{ }mol\text{ }NH_3*\frac{1\text{ }mol\text{ }H_2SO_4}{2\text{ }mol\text{ }NH_3}=0.0019024\text{ }H_2SO_4$

6) Molarity of H2SO4.

Moles: 0.0019024

Volume: 25.0 mL = 0.025 L

$M=\frac{0.0019024\text{ }mol\text{ }H_2SO_4}{0.025\text{ }L}=0.076096\text{ }M\text{ }H_2SO_4$

The concentration of H2SO4 is 0.08 mol/L.

Option D.

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