Question

For each of the following solutions, the mass of solute taken is indicated, as well as the total volume of solution prepared. Calculate the normality of each solution. a. \(15.0 \mathrm{~g}\) of \(\mathrm{HCl} ; 500 . \mathrm{mL}\) b. \(49.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4} ; 250 . \mathrm{mL}\) c. \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4} ; 100 . \mathrm{mL}\)

Short answer

The normalities of the solutions are: a. HCl: \(0.822 \mathrm{~N}\) b. H2SO4: \(4 \mathrm{~N}\) c. H3PO4: \(3.06 \mathrm{~N}\)

Step-by-step solution

Calculate the equivalent mass of the solutes

For HCl, the equivalent mass is equal to its molar mass, which is \(36.5 \mathrm{~g/mol}\). For H2SO4, the equivalent mass is half its molar mass, which is \(98.0 \mathrm{~g/mol}\). For H3PO4, the equivalent mass is a third of its molar mass, which is \(98.0 \mathrm{~g/mol}\).

Calculate the gram-equivalent for the solutes

For HCl, gram-equivalent = mass of solute / equivalent mass = \(15.0 \mathrm{~g}\) / \(36.5 \mathrm{~g/mol}\) = \(0.411 \mathrm{~eq}\). For H2SO4, gram-equivalent = mass of solute / equivalent mass = \(49.0 \mathrm{~g}\) / \(49.0 \mathrm{~g/mol}\) = \(1 \mathrm{~eq}\). For H3PO4, gram-equivalent = mass of solute / equivalent mass = \(10.0 \mathrm{~g}\) / \(32.67 \mathrm{~g/mol}\) = \(0.306 \mathrm{~eq}\).

Convert the volume of solutions to liters

For HCl, volume = \(500 \mathrm{~mL}\) = \(0.500 \mathrm{~L}\). For H2SO4, volume = \(250 \mathrm{~mL}\) = \(0.250 \mathrm{~L}\). For H3PO4, volume = \(100 \mathrm{~mL}\) = \(0.100 \mathrm{~L}\).

Calculate the normality

For HCl, normality (N) = gram-equivalent / volume = \(0.411 \mathrm{~eq}\) / \(0.500 \mathrm{~L}\) = \(0.822 \mathrm{~N}\). For H2SO4, normality (N) = gram-equivalent / volume = \(1 \mathrm{~eq}\) / \(0.250 \mathrm{~L}\) = \(4 \mathrm{~N}\). For H3PO4, normality (N) = gram-equivalent / volume = \(0.306 \mathrm{~eq}\) / \(0.100 \mathrm{~L}\) = \(3.06 \mathrm{~N}\). The normalities of the solutions are: a. HCl: \(0.822 \mathrm{~N}\) b. H2SO4: \(4 \mathrm{~N}\) c. H3PO4: \(3.06 \mathrm{~N}\)

Key concepts

Equivalent Mass

Equivalent mass is a key concept in chemistry, especially when dealing with chemical reactions and solution preparation. It refers to the mass of a substance that will react with or supply one mole of hydrogen ions (H+) in an acid–base reaction, or one mole of electrons in a redox reaction.

To understand this, consider hydrochloric acid (HCl). Each molecule can release one hydrogen ion; therefore, its equivalent mass is the same as the molar mass, which is 36.5 g/mol.

• For HCl, the equivalent mass is 36.5 g/mol.
• For sulfuric acid (H2SO4), which can release two hydrogen ions per molecule, the equivalent mass is half of its molar mass: 49 g/mol.
• For phosphoric acid (H3PO4), which can release three hydrogen ions, its equivalent mass is a third of its molar mass: 32.67 g/mol.

Understanding equivalent mass helps in calculating the gram-equivalent, which is crucial for determining solution normality.

Gram-Equivalent

Gram-equivalent is a measure that helps quantify the reactive capacity of a solute in a solution. It is calculated by dividing the mass of the solute by its equivalent mass. This concept is vital for determining the concentration of a solution in terms of normality.

Here's how you calculate it:

• For HCl: gra-equivalent = 15.0 g / 36.5 g/mol = 0.411 eq.
• For H2SO4: gram-equivalent = 49.0 g / 49.0 g/mol = 1 eq.
• For H3PO4: gram-equivalent = 10.0 g / 32.67 g/mol = 0.306 eq.

The gram-equivalent helps pinpoint exactly how much of the substance is available to react in a solution. This is a stepping stone to calculating a solution's normality, hence understanding the concentration of reactive species present.

Solution Concentration

Solution concentration in terms of normality (N) offers insight into how reactive a solution is. Normality is determined by the ratio of gram-equivalent of a solute to the volume of the solution in liters.

Finish by converting volumes to liters for all calculations:

• HCl: volume = 500 mL = 0.500 L.
• H2SO4: volume = 250 mL = 0.250 L.
• H3PO4: volume = 100 mL = 0.100 L.

Then, calculate normality:

• For HCl, normality = 0.411 eq / 0.500 L = 0.822 N.
• For H2SO4, normality = 1 eq / 0.250 L = 4 N.
• For H3PO4, normality = 0.306 eq / 0.100 L = 3.06 N.

These calculations highlight how concentration captures the potential chemical reactivity within a given volume of solution. Knowing the normality assists in preparing solutions for specific reactions, ensuring the accurate delivery of ions or reactive components.