Question

A solution is made by mixing \(1.5 \mathrm{~g}\) of \(\mathrm{LiOH}\) and \(23.5 \mathrm{~mL}\) of \(1.000 \mathrm{M}\) HNO3. (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 resulting solution acidic or basic?

Short answer

The resulting solution is basic with [Li⁺] and [OH⁻] both at 1.659 M.

Step-by-step solution

Write the Chemical Reaction Equation

To find the balanced chemical equation for the reaction between lithium hydroxide (LiOH) and nitric acid (HNO₃), note that LiOH is a strong base and HNO₃ is a strong acid. They react to form lithium nitrate (LiNO₃) and water (H₂O):\[\text{LiOH} + \text{HNO}_3 \rightarrow \text{LiNO}_3 + \text{H}_2\text{O}\]This equation is already balanced as each side contains one Li, one OH, one NO₃, and one H₂O.

Calculate Moles of Reactants

We first need to find the number of moles of LiOH and HNO₃. For LiOH, use the molar mass of LiOH (ffffffffffff24.0 ffffffffff8 g/mol) and calculate:\[\text{Moles of LiOH} = \frac{1.5 \text{ g}}{24.0 \text{ g/mol}} = 0.0625 \text{ mol}\]For HNO₃, since the concentration is 1.000 M and the volume is 23.5 mL (convert it to liters by dividing by 1000):\[\text{Moles of HNO}_3 = 1.000 \times 0.0235 \text{ L} = 0.0235 \text{ mol}\]

Determine the Limiting Reactant

Compare the moles of the reactants given in the reaction, where a 1:1 molar ratio is required. Since LiOH and HNO₃ react in a 1:1 ratio, and we have 0.0625 mol of LiOH and 0.0235 mol of HNO₃, HNO₃ is the limiting reactant.

Calculate Remaining Moles of Excess Reactant

Since HNO₃ is the limiting reactant, all 0.0235 mol of it will react:Subtract the moles of HNO₃ reacted from the moles of LiOH:\[\text{Remaining moles of LiOH} = 0.0625 - 0.0235 = 0.039 \text{ mol}\]Since LiOH is excess, this amount remains in solution.

Calculate Concentration of Ions in Solution

Calculate the concentration of remaining ions. Since all HNO₃ is consumed, [H⁺] = 0.Remaining [Li⁺] from remaining LiOH, assuming the final volume is the sum of volumes of solutions:Total volume = 23.5 mL (HNO₃) + volume of water from LiOH ≈ 23.5 mL (volume of solid negligible)Convert to liters: 0.0235 L\[\text{Concentration of Li}^+ = \frac{0.039 \text{ mol}}{0.0235 \text{ L}} = 1.659 \text{ M}\][OH⁻] from remaining LiOH is the same:\[[\text{OH}^-] = 1.659 \text{ M}\]

Determine Solution pH

Since OH⁻ ions are in excess, the resulting solution is basic.

Key concepts

Balanced Chemical Equation

When dealing with chemical reactions, balancing the chemical equation is crucial to represent the reaction accurately. This ensures that the same number of each type of atom appears on both sides of the equation, complying with the law of conservation of mass. In the reaction between lithium hydroxide (LiOH) and nitric acid (HNO₃), we form lithium nitrate (LiNO₃) and water (H₂O). The reaction can be expressed as follows:
\[\text{LiOH} + \text{HNO}_3 \rightarrow \text{LiNO}_3 + \text{H}_2\text{O}\]
This equation is balanced, meaning we have equal numbers of Li, OH, NO₃, and H atoms on both sides. Often, balancing involves adjusting coefficients in front of the compounds, but here, each compound has a coefficient of 1, making it straightforward.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, limiting the amount of product that can be formed. Identifying the limiting reactant is essential for calculating the maximum amount of product possible from a reaction.
To find the limiting reactant in this example, we need to calculate the number of moles of each reactant. We find that there are 0.0625 moles of LiOH and 0.0235 moles of HNO₃. Given the 1:1 molar ratio needed for the reaction, HNO₃ is the limiting reactant as it has fewer moles available than LiOH. Once all of the HNO₃ reacts, it limits the reaction from proceeding further and dictates the amount of product, LiNO₃, that can be formed.

Concentration Calculation

After the reaction completes, it’s important to determine the concentration of any remaining ions in the solution. Concentration is typically expressed in moles per liter (M). In this problem, the total volume of the solution after reaction is approximately 23.5 mL or 0.0235 L.
Since HNO₃ is the limiting reactant and completely consumed, there are no remaining H⁺ ions. However, there are leftover LiOH molecules, specifically 0.039 moles, which did not react with HNO₃. This leads to lithium ions (Li⁺) and hydroxide ions (OH⁻) remaining in solution. The concentration for each of these ions is:
\[\text{Concentration of } \text{Li}^+ = \frac{0.039 \text{ mol}}{0.0235 \text{ L}} = 1.659 \text{ M}\]
The value also applies to [OH⁻]. In this case, concentration calculations help us determine the new chemical milieu of the solution.

Solution pH Determination

The pH of a solution is an indication of how acidic or basic it is. A pH less than 7 is acidic, a pH of 7 is neutral, and a pH greater than 7 is basic. Since the reaction in our problem consumes all H⁺ ions and leaves a surplus of OH⁻ ions in the solution, the result is a basic solution.
The concentration of hydroxide ions, [OH⁻], directly impacts pH. We can find the pOH and then the pH of the solution. The pOH is calculated using:
\[\text{pOH} = -\log [\text{OH}^-] = -\log(1.659) \approx 0.780\]
From the relationship between pH and pOH:
\[\text{pH} + \text{pOH} = 14\]
We solve for pH:
\[\text{pH} = 14 - 0.780 = 13.220\]
This confirms the solution is indeed basic, as expected from the excess of OH⁻ ions.