Question

Zn can reduce \(\mathrm{NO}_{3}^{-}\) to \(\mathrm{NH}_{3}(\mathrm{g})\) in basic solution. (The following equation is not balanced.) $$\begin{aligned}\mathrm{NO}_{3}^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})+& \mathrm{OH}^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \\\&\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(\mathrm{aq})+\mathrm{NH}_{3}(\mathrm{g})\end{aligned}$$ The \(\mathrm{NH}_{3}\) can be neutralized with an excess of \(\mathrm{HCl}(\mathrm{aq})\) Then, the unreacted HCl can be titrated with NaOH. In this way a quantitative determination of \(\mathrm{NO}_{3}^{-}\) can be achieved. A 25.00 mL sample of nitrate solution was treated with zinc in basic solution. The \(\mathrm{NH}_{3}(\mathrm{g})\) was passed into \(50.00 \mathrm{mL}\) of \(0.1500 \mathrm{M} \mathrm{HCl} .\) The excess \(\mathrm{HCl}\) required \(32.10 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{NaOH}\) for its titration. What was the \(\left[\mathrm{NO}_{3}\right]\) in the original sample?

Short answer

The original concentration of \(NO_{3}^{-}\) in the sample was \(0.172 \, M\).

Step-by-step solution

Determine the moles of NaOH used in the titration of excess HCl

The volume of \(0.1000 \,M\) \(NaOH\) used was \(32.10\, mL\), or \(32.10 \times10^{-3}\, L\). The molarity (M) is defined as moles/L, which we can rearrange to find the moles of \(NaOH\) as \(M \times \text{Volume in L}\), \(0.1000 \times 32.10 \times10^{-3} = 3.210 \times 10^{-3}\, \text{moles of NaOH}\).

Determine the moles of HCl that reacted with NH3

The moles of \(NaOH\) and \(HCl\) will be the same because \(NaOH\) and \(HCl\) react in a 1:1 ratio. Therefore, there were \(3.210 \times 10^{-3}\, moles\) of \(HCl\) unused. The initial amount of \(HCl\) was \(50.00 \times 10^{-3} \, L \times 0.1500 \, M = 7.500 \times 10^{-3}\, \text{moles}\). Thus, the moles of \(HCl\) that reacted with \(NH_{3}\) is \(7.500 \times 10^{-3} - 3.210 \times 10^{-3} = 4.290 \times 10^{-3}\, \text{moles}\).

Determine the concentration of \(NO_{3}^{-}\) in the original sample

Since \(NO_{3}^{-}\) was converted into \(NH_{3}\) in a 1:1 ratio, the moles of \(NH_{3}\) is also \(4.290 \times 10^{-3}\, \text{moles}\). Lastly, divide the number of moles of nitrate by the volume of the original nitrate solution to get the original concentration, \(4.290 \times 10^{-3} \, moles / (25.00 \times 10^{-3}) \, L = 0.172 \, M\).

Key concepts

Redox Reactions

Redox reactions, also known as oxidation-reduction reactions, are chemical processes that involve the transfer of electrons between two substances. This kind of reaction is crucial in the exercise involving the reduction of nitrate ions, (\(\mathrm{NO}_3^-\)), by zinc metal (\(\mathrm{Zn}\)).
At its core, a redox reaction comprises two complementary processes: oxidation, where a substance loses electrons, and reduction, where another gains those electrons. In our scenario, zinc is oxidized by losing electrons, while the nitrate ions undergo reduction by gaining electrons to eventually form ammonia (\(\mathrm{NH}_3\)).

• The oxidizing agent causes the oxidation of another substance and is itself reduced. In our reaction, nitrate ions act as the oxidizing agent.
• The reducing agent gives away electrons, undergoing oxidation. Zinc acts as the reducing agent.

In basic solutions, these processes help facilitate the electron transfer process needed to convert \(\mathrm{NO}_3^-\) to \(\mathrm{NH}_3\) while producing zinc hydroxide complexes.

Titration

Titration is an analytical method used to determine the concentration of an unknown solution by reacting it with a standard solution of known concentration. In our example, the excess hydrochloric acid (\(\mathrm{HCl}\)) that did not react with ammonia is quantified using titration with sodium hydroxide (\(\mathrm{NaOH}\)).
The process is straightforward: \(\mathrm{NaOH}\) is added to the remaining \(\mathrm{HCl}\) until neutralization, indicated by a pH change or an appropriate color change in an indicator. The amount of \(\mathrm{NaOH}\) used reveals how much \(\mathrm{HCl}\) remained unreacted, allowing us to calculate the moles of \(\mathrm{HCl}\) that were involved in the actual reaction with \(\mathrm{NH}_3\).

• We begin with the initial known volume and concentration of \(\mathrm{HCl}\).
• By measuring the amount of \(\mathrm{NaOH}\) used to reach neutralization, we ascertain the exact amount of excess \(\mathrm{HCl}\).

This effective technique allows for quantitative determination of substances in solution, like the concentration of nitrate ions in this context.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves the calculation of reactants and products in chemical reactions. It plays a fundamental role in solving the given problem, as it provides the pathway to understanding how the reactants convert into products.
The stoichiometry of a reaction gives us the exact molar ratios in which reactants combine and products form. For example, in the redox reaction reducing nitrate ions to ammonia, each mole of nitrate forms one mole of ammonia. This 1:1 stoichiometric relationship is vital for understanding how much ammonia is produced.

• Stoichiometric coefficients are derived from balanced chemical equations.
• They enable calculations of moles from known quantities of one or more substances.

In practical terms, if you know the moles of a transformed compound, you can calculate the initial moles of the reactant, like \(\mathrm{NO}_3^-\), giving way to further determining its concentration.

Molarity Calculations

Molarity (\(\mathrm{M}\)) expresses the concentration of a solution as moles of solute per liter of solution. Calculating molarity is essential to understand how concentrated a solution is, which is pivotal when dealing with reactions where precise amounts are needed.
In the exercise at hand, to find the concentration of \(\mathrm{NO}_3^-\) ions in the original sample, it involves several straightforward calculations:

• First, calculate the moles of a product or reactant utilizing stoichiometry and known molarity values.
• Then, the moles of the substance are divided by the volume (in liters) of the solution where the substance is contained.

For instance, if you determine that the moles of \(\mathrm{NH}_3\) produced are known, and therefore the initial moles of \(\mathrm{NO}_{3}^{-}\) are also known (through stoichiometry), you proceed to divide this by the sample's volume to obtain molarity, reflecting the concentration in \(\mathrm{mol/L}\).