Calculate the volume of \(0.150 \mathrm{M} \mathrm{HCl}(\mathrm{aq})\) required to neutralize (a) one-half and (b) all the hydroxide ions in \(25.0 \mathrm{~mL}\) of \(0.110 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})\). (c) What is the molarity of \(\mathrm{Na}^{*}\) ions at the stoichiometric point? (d) Calculate the \(\mathrm{pH}\) of the solution after the addition of \(20.0 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) HClaq) to \(25.0 \mathrm{~mL}\) of \(0.110 \mathrm{M} \mathrm{NaOH}(\mathrm{aq}) .\)

Stoichiometry refers to the calculation of reactants and products in chemical reactions. In the context of neutralization reactions, stoichiometry involves balancing the number of acid and base moles that are required to achieve a neutral solution, which is a solution with equal numbers of hydrogen ions (\text{H}^+) and hydroxide ions (\text{OH}^-). The stoichiometric point is where the amount of acid equals the amount of base according to their balanced equation.

For instance, the neutralization reaction between hydrochloric acid (\text{HCl}) and sodium hydroxide (\text{NaOH}) can be represented as follows:
\[\[\begin{align*} \text{HCl (aq)} + \text{NaOH (aq)} &\rightarrow \text{NaCl (aq)} + \text{H}_2\text{O (l)} \end{align*}\]\] In this reaction, one mole of \text{HCl} reacts with one mole of \text{NaOH} to produce one mole of \text{NaCl} and one mole of water, \text{H}_2\text{O}. When solving stoichiometry problems, it is crucial to start with a balanced chemical equation and use molarity and volume to find the moles of reactants. From there, you can find out how much of the other reactant is needed to complete the reaction.

Molarity (\text{M}) is a unit of concentration that measures the number of moles of a solute per liter of solution. To understand neutralization, you must be able to calculate the molarity of acids and bases. This involves using the formula: \[\[\begin{align*} \text{Molarity} &= \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \end{align*}\]\] In practice, if you know the molarity of a solution and the volume of the solution you're using, you can calculate the number of moles of the solute that are present. For example, when you're given the volume of \text{NaOH} solution and its molarity, you can find the number of moles of \text{NaOH}:

\[\[\begin{align*} \text{Moles of NaOH} &= \text{Molarity of NaOH} \times \text{Volume of NaOH in liters} \end{align*}\]\] In neutralization, you'll need to use molarity to determine how much acid is needed to neutralize a given amount of base, or vice versa, which is directly linked to the stoichiometry of the reaction.

pH is a scale that measures the acidity or basicity of an aqueous solution. The pH scale ranges from 0 to 14, where 7 is neutral, values less than 7 are acidic, and values greater than 7 are basic. pH can be calculated using the concentration of hydrogen ions (\text{H}^+) in the solution:

\[\[\begin{align*} \text{pH} &= -\text{log}_{10} [\text{H}^+] \end{align*}\]\] For instance, to find the pH of an excess acid after neutralization, you would find the moles of the excess acid, calculate its molarity in the new solution volume, and then take the negative log of that concentration.

It is important for students to understand how to conduct these calculations as they often come across them when dealing with acid-base reactions. In the case of a strong acid like \text{HCl}, the concentration of \text{H}^+ is equal to the molarity of \text{HCl} because it dissociates completely in water. This concept becomes particularly essential after a neutralization reaction, which can change the pH of the solution. By mastering pH calculations, students can predict the acidity or basicity of the resultant solution.

Acid-base reactions are chemical reactions that occur between acids and bases. These reactions usually result in the formation of water and a salt. Acids are substances that can donate a hydrogen ion (\text{H}^+), while bases are substances that can accept a hydrogen ion. When an acid reacts with a base, the hydrogen ions from the acid combine with the hydroxide ions (\text{OH}^-) from the base to form water (\text{H}_2\text{O}).

Understanding acid-base reactions is crucial for working on neutralization problems since these reactions are the ones being balanced and calculated. The general equation \[\[\begin{align*} \text{Acid} + \text{Base} &\rightarrow \text{Salt} + \text{H}_2\text{O} \end{align*}\]\] exemplifies this type of reaction. Students must grasp this concept to comprehend how stoichiometry, molarity, and pH calculations relate to each other during neutralization. Moreover, recognizing that neutralizations involve equal quantities of acid and base reacting, and that the products have different properties from the reactants, is key to solving these problems efficiently.