A 500.-mL solution consists of \(0.050 \mathrm{~mol}\) of solid \(\mathrm{NaOH}\) and \(0.13 \mathrm{~mol}\) of hypochlorous acid \(\left(\mathrm{HClO} ; K_{\mathrm{a}}=3.0 \times 10^{-8}\right)\) dissolved in water. (a) Aside from water, what is the concentration of each species that is present? (b) What is the \(\mathrm{pH}\) of the solution? (c) What is the pH after adding \(0.0050 \mathrm{~mol}\) of \(\mathrm{HCl}\) to the flask?

A buffer solution is a system that resists changes in pH upon the addition of small amounts of acid or base. It typically contains a weak acid and its conjugate base or a weak base and its conjugate acid.

To understand how a buffer works, we use the buffer equation, also known as the Henderson-Hasselbalch equation: \[ \text{pH} = pK_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \] This equation allows us to calculate the pH of a buffer solution based on the concentrations of the acid (\text{HA}) and its conjugate base (\text{A}^-), along with the acid dissociation constant (\text{pK}_a).

The Henderson-Hasselbalch equation is a derived equation from the expression of the ionization constant (\text{K}_a). It relates the pH of a solution to the pK_a (negative log of \text{K}_a) and the ratio of the concentration of the deprotonated form to the protonated form of the acid: \[ \text{pH} = pK_a + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \]

This equation is particularly useful for calculating the pH of buffer solutions. In the given exercise, the acid (\text{HClO}) and its conjugate base (\text{ClO}^-) form the buffer system. By knowing the concentrations of these species and the \text{pK}_a, we can apply the Henderson-Hasselbalch equation to find the solution's pH.

The ionization constant (\text{K}_a) is a measure of the strength of an acid in solution. For the weak acid hypochlorous acid (\text{HClO}), the ionization reaction in water is: \ \[ \text{HClO} \rightleftharpoons \text{H}^+ + \text{ClO}^- \] \[ \text{K}_a = \frac{[\text{H}^+][\text{ClO}^-]}{[\text{HClO}]} \]

Given \text{K}_a value helps us understand how much the acid dissociates in water. A smaller \text{K}_a value, like 3.0 \times 10^{-8} for \text{HClO}, indicates a weak acid that doesn't dissociate completely, leading to small concentrations of \text{H}^+ and \text{ClO}^- ions.

A strong base, such as sodium hydroxide (\text{NaOH}), dissociates completely in water to give \text{OH}^- ions: \ \[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

Because the strong base fully dissociates, the concentration of \text{OH}^- ions directly equates to the concentration of the \text{NaOH} solution. In the exercise, \text{NaOH} reacts with the \text{HClO}, converting it to \text{ClO}^-, which impacts the buffer components and ultimately the solution's pH.

A neutralization reaction occurs when an acid and a base react to form water and a salt. In the context of buffers, adding a strong acid (\text{HCl}) to the buffer system impacts the pH in the following way: \ \[ \text{HCl} + \text{ClO}^- \rightarrow \text{HClO} + \text{Cl}^- \]

When hydrochloric acid (\text{HCl}) is added to the buffer solution described in the exercise, the \text{H}^+ ions from the \text{HCl} neutralize the \text{ClO}^- ions to form more \text{HClO}. This shifts the concentrations of the species within the buffer, and applying the Henderson-Hasselbalch equation again allows us to find the new pH value, demonstrating the buffer's ability to resist drastic changes in pH.