Question

Calculate the normality of each of the following solutions. a. \(0.134 \mathrm{M} \mathrm{NaOH}\) b. \(0.00521 M \mathrm{Ca}(\mathrm{OH})_{2}\) c. \(4.42 M \mathrm{H}_{3} \mathrm{PO}_{4}\)

Short answer

The normality of the given solutions are: a. \(0.134 N\, NaOH\) b. \(0.01042 N\, Ca(OH)_2\) c. \(13.26 N\, H_3PO_4\)

Step-by-step solution

Identify the number of replaceable ions

For each solute, we need to identify the number of replaceable hydrogen ions (acids) or hydroxide ions (bases): a. NaOH: For every molecule of NaOH, there is 1 hydroxide ion (\(OH^-\)). Hence, the number of replaceable ions per molecule of NaOH = 1. b. Ca(OH)₂: For every molecule of Ca(OH)₂, there are 2 hydroxide ions (\(OH^-\)). Hence, the number of replaceable ions per molecule of Ca(OH)₂ = 2. c. H₃PO₄: For every molecule of H₃PO₄, there are 3 hydrogen ions (\(H^+\)). Hence, the number of replaceable ions per molecule of H₃PO₄ = 3.

Calculate the normality of each solution

Using the formula N = M × number of replaceable ions, we can calculate the normality for each solution: a. For the \(0.134 M \, NaOH\) solution: Normality (N) = Molarity (M) × number of replaceable ions = \(0.134 M \times 1 = 0.134 N\) b. For the \(0.00521 M \, Ca(OH)_2\) solution: Normality (N) = Molarity (M) × number of replaceable ions = \(0.00521 M \times 2 = 0.01042 N\) c. For the \(4.42 M \, H_3PO_4\) solution: Normality (N) = Molarity (M) × number of replaceable ions = \(4.42 M \times 3 = 13.26 N\) Thus, the normality of the given solutions are: a. \(0.134 N\, NaOH\) b. \(0.01042 N\, Ca(OH)_2\) c. \(13.26 N\, H_3PO_4\)

Key concepts

Normality of Solutions

The concept of normality is a way to express concentration that is particularly useful in reactions involving acids and bases. Normality (N) is defined as the number of equivalents of solute per liter of solution. An equivalent, in this context, is the amount of a substance that reacts with or is equivalent to an arbitrary amount of another substance in a given chemical reaction.

Normality can be calculated using the following formula: \[ N = M \times n \] Here, \(N\) is the normality, \(M\) is the molarity, and \(n\) is the number of equivalents per mole of solute, which is often the same as the number of replaceable hydrogen ions in an acid or the number of hydroxide ions in a base. For instance, hydrochloric acid (HCl) has one replaceable hydrogen ion, so one mole of HCl is equivalent to one equivalent. Understanding normality helps in stoichiometric calculations, especially in titration problems.

Molarity to Normality Conversion

Converting molarity to normality is a crucial step in many chemical calculations and can be easily done using the number of equivalents. Since molarity (M) is based on moles of solute per liter of solution and normality (N) is based on equivalents per liter, the relationship between the two can be established by multiplying the molarity by the number of equivalents (n).

For example, in our exercise, sodium hydroxide (NaOH) has a molarity of \(0.134 M\). Since NaOH has one hydroxide ion which can disassociate, it has one equivalent per mole. The normality would thus be equal to the molarity:
\[ N = M \times n = 0.134 M \times 1 = 0.134 N \]
On the other hand, calcium hydroxide (\(Ca(OH)_2\)) has two hydroxide ions per molecule, making \(n=2\). Therefore, the normality is double the molarity:
\[ N = M \times 2 \]

This conversion is an excellent example of how understanding these relationships can help simplify complex problems.

Equivalent Concept in Chemistry

The equivalent concept in chemistry refers to the use of equivalents to make comparisons between reactions and reactants. An equivalent is a measure that indicates the reactive capacity of a molecule and depends on the context of the chemical reaction involved. It is integral to the idea of normality.

Equivalents are especially useful when dealing with acid-base reactions, redox reactions, and precipitation reactions. Since different molecules react in different proportions, the 'equivalent' helps to even out these differences and allows for the direct comparison of substances. For acids, an equivalent corresponds to one mole of hydrogen ions (\(H^+\)) that can be donated. For bases, it equates to one mole of hydroxide ions (\(OH^-\)) that can be accepted.

As seen in our exercise, for each molecule of \(H_3PO_4\), there are three hydrogen ions, hence three equivalents. This implies that for the purposes of reactions involving \(H_3PO_4\), each mole of \(H_3PO_4\) is equivalent to three moles of a monoprotic acid, such as HCl. The understanding of equivalents is therefore pivotal to accurately calculating normality and performing stoichiometric calculations.