Question

A solution is made by mixing \(12.0 \mathrm{~g}\) of \(\mathrm{NaOH}\) and \(75.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HNO}_{3}\). (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution. (c) Is the resultant solution acidic or basic?

Short answer

The balanced equation for the reaction between $\mathrm{NaOH}$ and $\mathrm{HNO_3}$ is: \( NaOH + HNO_3 \rightarrow NaNO_3 + H_2O \). The concentration of $\mathrm{OH^-}$ remaining in the solution is \(0 \mathrm{M}\), and the concentration of $\mathrm{H^+}$ remaining is \(0.00020 \mathrm{M}\). Since the concentration of $\mathrm{H^+}$ ions is greater than the $\mathrm{OH^-}$ ions, the resultant solution is acidic.

Step-by-step solution

Balanced equation for the reaction

The reaction between NaOH and HNO3 is a neutralization reaction, with NaOH being a strong base and HNO3 being a strong acid. The reaction produces water and a salt. The balanced equation for this reaction is: \( NaOH + HNO_3 \rightarrow NaNO_3 + H_2O \)

Calculate concentration of each ion before the reaction

We are given that the mass of NaOH is 12.0 g and the volume of HNO3 solution is 75.0 mL. We can use these volumes and masses to calculate the initial concentration of each ion. a) Convert the mass of NaOH to moles. - molar mass of NaOH: \( 23 + 16 + 1 = 40 \mathrm{~g/mol} \) - moles: \( \dfrac{12.0 \mathrm{~g}}{40 \mathrm{~g/mol}} = 0.300 \mathrm{~mol} \) b) Calculate the total volume of the solution. - Since the volume of the NaOH is not given, we assume it is negligible compared to the 75.0 mL of HNO3 solution. Therefore, we will use 75.0 mL as the total volume of the solution. c) Calculate the concentration of each ion in the solution before the reaction. - concentration of OH- from NaOH: \( \dfrac{0.300 \mathrm{~mol}}{75.0 \mathrm{~mL}} = 0.004 \mathrm{M} \) - concentration of H+ from HNO3: \( 0.200 \mathrm{M} \)

Determine the concentrations of each ion after the reaction

As NaOH and HNO3 react in a 1:1 ratio, the moles of H+ and OH- ions will cancel each other out. We will now subtract the moles of the reactants that were used up in the reaction to determine the remaining concentrations of each ion. a) Calculate the moles of OH- and H+ ions used up in the reaction. - moles of OH- used: \(0.300 \mathrm{~mol} \) - moles of H+ used: \( 0.200 \mathrm{M} \times 75.0 \mathrm{~mL} = 0.015 \mathrm{~mol} \) b) Calculate the concentration of each ion remaining in solution. - concentration of OH- remaining: \( 0.004 \mathrm{M} - \dfrac{0.300 \mathrm{~mol}}{75.0 \mathrm{~mL}} = 0 \mathrm{M} \) - concentration of H+ remaining: \( 0.200 \mathrm{M} - \dfrac{0.015 \mathrm{~mol}}{75.0 \mathrm{~mL}} = 0.00020 \mathrm{M} \)

Determine if the solution is acidic or basic

Based on the concentration of the remaining H+ and OH- ions in the solution, the concentration of the H+ ions is greater than the OH- ions (0.00020 M versus 0 M). Therefore, the resultant solution is acidic.

Key concepts

Neutralization Reaction

When an acid reacts with a base, they form a neutralization reaction, which produces water and a salt. This is a key concept in chemistry, as it helps in understanding how acids and bases interact to neutralize each other. In this context, the reaction between sodium hydroxide (\(\mathrm{NaOH}\)) and nitric acid (\(\mathrm{HNO_3}\)) serves as a classic example. Here, \(\mathrm{NaOH}\) is a strong base, and \(\mathrm{HNO_3}\) is a strong acid. When these two react:

• Sodium ions (\(\mathrm{Na^+}\)) combine with nitrate ions (\(\mathrm{NO_3^-}\)) to form sodium nitrate (\(\mathrm{NaNO_3}\)), a soluble salt.
• Hydroxide ions (\(\mathrm{OH^-}\)) and hydrogen ions (\(\mathrm{H^+}\)) combine to form water (\(\mathrm{H_2O}\)).

The equation for this neutralization is balanced as follows: \[\mathrm{NaOH} + \mathrm{HNO_3} \rightarrow \mathrm{NaNO_3} + \mathrm{H_2O}\] This reaction demonstrates the principle of conservation of mass, as the number of each type of atom is the same before and after the reaction.

Ion Concentration Calculation

Calculating ion concentrations in a solution helps in understanding the stoichiometry and extent of reactions between solutes. Initially, we need to determine the amounts of each reactant in terms of moles. For \(\mathrm{NaOH}\), the molar mass is \(40\) g/mol, which means \(12\) grams converts to \(0.300\) moles. The \(\mathrm{HNO_3}\) is provided as a \(0.200 \, \mathrm{M}\) solution in \(75.0 \, \mathrm{mL}\), equating to \(0.015\) moles. For the reaction, \(\mathrm{OH^-}\) and \(\mathrm{H^+}\) ions will react in a 1:1 ratio. This means:

• All \(\mathrm{0.015}\) moles of \(\mathrm{H^+}\) from \(\mathrm{HNO_3}\) will be consumed.
• For \(\mathrm{NaOH}\), the \(\mathrm{0.300}\) moles is excess, with only \(\mathrm{0.015}\) moles reacting.

The remaining concentration of ions can be determined based on the excess \(\mathrm{NaOH}\), leading to a concentration of zero for \(\mathrm{OH^-}\) and \(0.00020 \, \mathrm{M}\) for \(\mathrm{H^+}\) in the resultant solution.

Solution Acidity or Basicity

Once we know the concentration of hydrogen and hydroxide ions, we can determine if the solution is acidic, neutral, or basic. Acidity or basicity is often indicated by comparing the concentration of \(\mathrm{H^+}\) to \(\mathrm{OH^-}\) ions in a solution. In this case:

• There are zero moles per liter (\(\mathrm{0 \, M}\)) of hydroxide ions because all of them have reacted.
• The solution contains \(0.00020 \, \mathrm{M}\) of hydrogen ions.

Since the concentration of \(\mathrm{H^+}\) ions in the solution is greater than the concentration of \(\mathrm{OH^-}\) ions, the resultant solution is classified as acidic. In acidic solutions, the \(\mathrm{pH}\) value is less than 7. This understanding helps to determine the use or treatment of the solution in various chemical processes, and it reinforces the basic principles behind the pH scale.