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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 (HNO鈧�), we form lithium nitrate (LiNO鈧�) and water (H鈧侽). 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, NO鈧�, 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.

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 HNO鈧�. Given the 1:1 molar ratio needed for the reaction, HNO鈧� is the limiting reactant as it has fewer moles available than LiOH. Once all of the HNO鈧� reacts, it limits the reaction from proceeding further and dictates the amount of product, LiNO鈧�, that can be formed.

After the reaction completes, it鈥檚 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 HNO鈧� is the limiting reactant and completely consumed, there are no remaining H鈦� ions. However, there are leftover LiOH molecules, specifically 0.039 moles, which did not react with HNO鈧�. This leads to lithium ions (Li鈦�) and hydroxide ions (OH鈦�) 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鈦籡. In this case, concentration calculations help us determine the new chemical milieu of the solution.

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鈦� ions and leaves a surplus of OH鈦� ions in the solution, the result is a basic solution.
The concentration of hydroxide ions, [OH鈦籡, 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鈦� ions.