Question

Calculate the concentration of the acid (or base) remaining in solution when \(10.7 \mathrm{~mL}\) of \(0.211 \mathrm{M} \mathrm{HNO}_{3}\) is added to \(16.3 \mathrm{~mL}\) of \(0.258 \mathrm{M} \mathrm{NaOH}\).

Short answer

The remaining concentration of NaOH is 0.0721 M.

Step-by-step solution

Calculate Moles of HNO3

First, calculate the moles of the nitric acid, \(\mathrm{HNO}_3\), using the formula \( \text{moles} = M \times V \). Here, \(M = 0.211 \mathrm{~M}\) and \(V = 10.7 \times 10^{-3} \mathrm{~L}\). Thus, the moles of \(\mathrm{HNO}_3\) = \(0.211 \times 10.7 \times 10^{-3} = 0.0022577\) moles.

Calculate Moles of NaOH

Next, calculate the moles of the sodium hydroxide, \(\mathrm{NaOH}\), using the same formula. Here, \(M = 0.258 \mathrm{~M}\) and \(V = 16.3 \times 10^{-3} \mathrm{~L}\). Thus, the moles of \(\mathrm{NaOH}\) = \(0.258 \times 16.3 \times 10^{-3} = 0.0042054\) moles.

Determine Limiting Reactant

Compare the moles of \(\mathrm{HNO}_3\) and \(\mathrm{NaOH}\). The reaction is \(\mathrm{HNO}_3 + \mathrm{NaOH} \rightarrow \mathrm{NaNO}_3 + \mathrm{H}_2\mathrm{O}\). Since \(0.0022577\) moles of \(\mathrm{HNO}_3\) is less than \(0.0042054\) moles of \(\mathrm{NaOH}\), \(\mathrm{HNO}_3\) is the limiting reactant.

Calculate Remaining Moles of NaOH

The remaining moles of \(\mathrm{NaOH}\) are \(0.0042054 - 0.0022577 = 0.0019477\) moles, because \(\mathrm{NaOH}\) is in excess.

Calculate Total Volume

Calculate the total volume of the solution by adding the volumes of \(\mathrm{HNO}_3\) and \(\mathrm{NaOH}\): \(10.7 \mathrm{~mL} + 16.3 \mathrm{~mL} = 27.0 \mathrm{~mL} = 27.0 \times 10^{-3} \mathrm{~L}\).

Calculate Final Concentration of NaOH

Finally, calculate the concentration of the remaining \(\mathrm{NaOH}\) using \( \text{Concentration} = \frac{\text{moles}}{\text{volume}} \). Thus, the concentration of \(\mathrm{NaOH}\) is \(\frac{0.0019477}{27.0 \times 10^{-3}} = 0.0721 \mathrm{~M}\).

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that will be fully consumed first, thus halting the reaction. It's crucial to identify which reactant runs out first because this determines how much of each product can be formed. In our case with \(\mathrm{HNO}_3\) and \(\mathrm{NaOH}\), comparing the moles of each helps us find the limiting reactant. We calculated 0.0022577 moles of \(\mathrm{HNO}_3\) and 0.0042054 moles of \(\mathrm{NaOH}\), leading us to find that \(\mathrm{HNO}_3\) is the limiting reactant since there is less of it relative to the stoichiometric ratios of the reaction.
This concept means that all of the \(\mathrm{HNO}_3\) will react with some portion of the \(\mathrm{NaOH}\), and any excess \(\mathrm{NaOH}\) will remain unreacted. Understanding the role of the limiting reactant simplifies predicting the amounts of products formed and leftovers.

Molarity

Molarity, or molar concentration, measures how much of a substance is dissolved in a certain volume of solution. It's expressed as moles per liter (\(\mathrm{mol}/\mathrm{L}\) or \(\mathrm{M}\)). To find molarity, divide the number of moles of the solute by the solution's volume in liters.
In the given reaction, the initial molarity of \(\mathrm{HNO}_3\) is 0.211 \(\mathrm{M}\), calculated over a volume of 10.7 \(\mathrm{mL}\). For \(\mathrm{NaOH}\), it's 0.258 \(\mathrm{M}\) with a volume of 16.3 \(\mathrm{mL}\). Molarity helps us determine how concentrated each reactant is, thus aiding in finding the limiting reactant and predicting reaction outcomes. After the reaction, we find the concentration of any remaining reactant to know its molarity in the final mixture.

Stoichiometry

Stoichiometry involves calculating the relative quantities of reactants and products in chemical reactions. It means using the balanced chemical equation to find proportions between moles of reactants and products.
For this reaction, the balanced equation \(\mathrm{HNO}_3 + \mathrm{NaOH} \rightarrow \mathrm{NaNO}_3 + \mathrm{H}_2\mathrm{O}\) shows a 1:1 ratio between \(\mathrm{HNO}_3\) and \(\mathrm{NaOH}\). Stoichiometry tells us that every mole of \(\mathrm{HNO}_3\) will react with exactly one mole of \(\mathrm{NaOH}\) to produce products.

• The moles of \(\mathrm{HNO}_3\) initially present determine the maximum amount of \(\mathrm{NaOH}\) that can react, due to the 1:1 ratio.
• Using stoichiometry, we also calculate the number of leftover moles when one reactant is in excess, as seen with \(\mathrm{NaOH}\) in this example.
• Finally, stoichiometry enables us to find the concentration of the leftover reactant by dividing its moles by the total volume of the solution.

Through stoichiometry, you can effectively analyze and predict how a chemical reaction proceeds and determine the quantities of products and unreacted materials.