Question

For each solution, calculate the initial and final pH after the addition of 0.010 mol of \(\mathrm{HCl} .\) a. 500.0 \(\mathrm{mL}\) of pure water b. 500.0 \(\mathrm{mL}\) of a buffer solution that is 0.125 \(\mathrm{M}\) in \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and 0.115 \(\mathrm{M}\) in \(\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) c. 500.0 \(\mathrm{mL}\) of a buffer solution that is 0.155 \(\mathrm{M}\) in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) and 0.145 \(\mathrm{M}\) in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\)

Short answer

For pure water, initial pH is 7 and final pH will be lower due to addition of HCl. For the acetic acid-sodium acetate buffer, initial pH can be calculated with the Henderson-Hasselbalch equation and the final pH changes slightly after adding HCl. For the ethylamine buffer, initial pH is calculated similarly and the final pH also changes slightly after HCl addition.

Step-by-step solution

- Finding the initial pH of pure water

Since pure water is neutral, its initial concentration of hydrogen ions is equal to its concentration of hydroxide ions, which is calculated using Kw, the ion-product constant for water. Kw is equal to 1.0 x 10^(-14) at 25°C, and the concentration of hydrogen ions is the square root of Kw. Using pH = -log[H+], we find that the initial pH of water is 7.

- Calculating the final pH after adding HCl to water

To find the final pH of the solution, calculate the new concentration of H+ ions. Since HCl is a strong acid, it will dissociate completely in water. 0.010 mol HCl in 500 mL of water results in a concentration of 0.010 mol / 0.50 L = 0.020 M. Using pH = -log[H+], plug in the concentration to find the pH.

- Initial pH of the acetic acid-sodium acetate buffer

To find the initial pH of the buffer solution, use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is for acetic acid (4.76), [A-] is the concentration of sodium acetate (0.115 M), and [HA] is the concentration of acetic acid (0.125 M).

- Final pH after adding HCl to the buffer

The addition of HCl to the buffer reacts with the sodium acetate to form acetic acid. The change in concentration of acetic acid and sodium acetate should be calculated. Then use the Henderson-Hasselbalch equation with the new concentrations to find the final pH.

- Initial pH of the ethylamine-ethylammonium chloride buffer

For the ethylamine buffer, use the Henderson-Hasselbalch equation for basic solutions: pOH = pKb + log([A-]/[HB+]), where [A-] is the concentration of ethylamine (0.155 M) and [HB+] is the concentration of ethylammonium chloride (0.145 M). pKb of ethylamine is known, and pOH can be converted to pH.

- Final pH after adding HCl to the ethylamine buffer

The HCl added will react with the ethylamine and increase the concentration of ethylammonium ions. Calculate the new concentrations like in step 4. Then use the Henderson-Hasselbalch equation for basic solutions with these new concentrations to find the final pOH and convert it to pH.

Key concepts

Acid-Base Buffers

Understanding acid-base buffers is crucial when you're dealing with changes in pH levels. Buffers are mixtures usually consisting of a weak acid and its conjugate base. Their main role is to resist changes in pH when small amounts of an acid or a base are added to the solution. This is because the components of the buffer can 'absorb' the extra H+ or OH- ions without significantly altering the overall pH.

For example, when HCl is added to a buffer solution, the H+ ions from the HCl will interact with the weak base part of the buffer. In the exercise, for the acetic acid-sodium acetate buffer, HCl reacts with sodium acetate forming more acetic acid, hence, only slightly changing the pH. This is due to the fact that acetic acid is a weak acid and will not release a significant amount of H+ ions to the solution. Buffer solutions are widely used in biological systems and laboratory settings where maintaining a stable pH is essential.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an incredibly useful tool for understanding buffer solutions. It allows us to calculate the pH of a buffer system when we know the concentrations of an acid and its conjugate base, or vice versa. The equation is given by:

\[\begin{equation} pH = pKa + \text{log}\bigg(\frac{[A^-]}{[HA]}\bigg)\end{equation}\]
where \( pKa \) represents the acid dissociation constant (a measure of the strength of the acid), \([A^-]\) is the concentration of the conjugate base, and \([HA]\) is the concentration of the acid. In the context of our exercise, it's used to determine the pH of acetic acid-sodium acetate and ethylamine-ethylammonium chloride buffer solutions before and after the addition of HCl. By utilizing this equation, it becomes clear how adding a strong acid like HCl affects the ratio of \([A^-]/[HA]\) and, consequently, the pH of the buffer solution.

Strong Acid Behavior

Strong acids are characterized by their complete dissociation in water, essentially releasing all their hydrogen (H+) ions. This contrasts with weak acids, which only partially dissociate. Due to this complete dissociation, strong acids can significantly alter the pH of a solution, unlike an acid in a buffer system that resists changes in pH. In our exercise, HCl is the strong acid added to the different solutions.

The behavior of strong acids is crucial in calculating pH changes in part (a) of the exercise, where HCl is added to pure water. As HCl completely dissociates, we directly use its concentration to determine the final pH of the solution. Learning the distinction between strong and weak acids will form a solid foundation for understanding how they influence the pH of solutions.

pH and pOH Relationships

pH and pOH are critical concepts in acid-base chemistry. They are logarithmic measures of the concentration of hydrogen ions (\( [H^+] \)) and hydroxide ions (\( [OH^-] \)), respectively. The pH scale typically ranges from 0 (very acidic) to 14 (very basic), with 7 being neutral.

pH and pOH are inversely related and their sum is constant at 25℃ (\( pH + pOH = 14 \)). Whenever you know one of these values, you can easily find the other. For instance, in part (c) of our exercise, after calculating the pOH of the ethylamine buffer solution, we can determine the pH by simply subtracting the pOH from 14. This relationship is an indispensable part of acid-base chemistry, and grasping it allows students to confidently navigate between acidic and basic solutions and their respective calculations.