Question

Calculate the \(\mathrm{pH}\) of (a) a \(0.5 \mathrm{M}\) solution with respect to \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa} ;\) (b) the same solution after \(0.1\) mole \(\mathrm{HCl}\) per liter has been added to it. Assume that the volume is unchanged. \(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5}\).

Short answer

The initial pH of the \(0.5 \mathrm{M}\) solution of \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\) is approximately 5.06. After adding \(0.1\) mole of \(\mathrm{HCl}\) per liter, the pH decreases to approximately 4.98.

Step-by-step solution

Writing the equilibrium reaction

Write the equilibrium reaction for the acid-base system, considering \(\mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid) as the acid and its conjugate base, which is the ion \(\mathrm{CH}_{3} \mathrm{COO^-}\) (acetate ion). The reaction is: \(\mathrm{CH}_{3} \mathrm{COOH + H_{2}O \rightleftharpoons CH_{3}COO^{-} + H_{3}O^{+}}\)

Set up the ICE table

Setup the ICE (Initial-Change-Equilibrium) table for this reaction. The given concentration of the solution is \(0.5 \mathrm{M}\) for both \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\). Initial concentrations: \([\mathrm{CH}_{3} \mathrm{COOH}] = 0.5 \mathrm{M}\) \([\mathrm{CH}_{3} \mathrm{COO^{-}}] = 0.5 \mathrm{M}\) \([\mathrm{H}_{3} \mathrm{O^{+}}] = 0\) Change in concentrations: \([-x]\) \([+x]\) \([+x]\) Equilibrium concentrations: \([0.5-x]\) \([0.5+x]\) \([x]\)

Substitute the equilibrium concentrations into \(\mathrm{K_{a}}\) equation

Using the equilibrium concentrations from the ICE table, substitute these concentrations into the \(\mathrm{K_{a}}\) equation: \(\mathrm{K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}} = \frac{x(0.5+x)}{0.5-x}\)

Solve for \([H^{+}]\) and calculate pH

Using the given \(\mathrm{K_{a}} = 1.75 \times 10^{-5}\), solve for \(x\) (the equilibrium concentration of \([\mathrm{H_{3} \mathrm{O^{+}}}]\)): \(1.75 \times 10^{-5} = \frac{x(0.5+x)}{0.5-x}\) In this case, since \(\mathrm{K_{a}}\) is very small, the change in concentration (\(x\)) is negligible compared to the initial concentrations, so: \(1.75 \times 10^{-5} \approx \frac{x(0.5)}{0.5}\) Solve for \(x\): \(x \approx 8.75 \times 10^{-6} \mathrm{M}\) Now, calculate the pH using: \(\mathrm{pH = -\log[H^{+}]}\) \(\mathrm{pH} = -\log(8.75 \times 10^{-6}) \approx 5.06\) #a) The initial pH is approximately 5.06# #b) Calculating the pH after adding 0.1 mole of HCl per liter#

Adjust the concentrations of ions after adding HCl

After adding \(0.1\) mole \(\mathrm{HCl}\) per liter to the solution, adjust the concentrations of the ions. The added \(\mathrm{HCl}\) will react with the base \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) to produce \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{Cl^{-}}\): \([\mathrm{CH}_{3} \mathrm{COOH}] = 0.5 + 0.1 = 0.6 \mathrm{M}\) \([\mathrm{CH}_{3} \mathrm{COO^{-}}] = 0.5 - 0.1 = 0.4 \mathrm{M}\)

Set up a new ICE table and substitute the equilibrium concentrations into \(\mathrm{K_{a}}\) equation

Setup a new ICE table for this reaction with the adjusted concentrations. Initial concentrations: \([\mathrm{CH}_{3} \mathrm{COOH}] = 0.6 \mathrm{M}\) \([\mathrm{CH}_{3} \mathrm{COO^{-}}] = 0.4 \mathrm{M}\) \([\mathrm{H}_{3} \mathrm{O^{+}}] = 0\) Change in concentrations: \([-x]\) \([+x]\) \([+x]\) Equilibrium concentrations: \([0.6-x]\) \([0.4+x]\) \([x]\) Using the equilibrium concentrations from this new ICE table, substitute these concentrations into the \(\mathrm{K}_{a}}\) equation: \(\mathrm{K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}} = \frac{x(0.4+x)}{0.6-x}\)

Solve for \([H^{+}]\) and calculate the new pH

Using the given \(\mathrm{K_{a}} = 1.75 \times 10^{-5}\), solve for \(x\) (the equilibrium concentration of \([\mathrm{H_{3} \mathrm{O^{+}}}]\)): \(1.75 \times 10^{-5} = \frac{x(0.4+x)}{0.6-x}\) Again, since \(\mathrm{K_{a}}\) is very small, the change in concentration (\(x\)) is negligible compared to the initial concentrations, so: \(1.75 \times 10^{-5} \approx \frac{x(0.4)}{0.6}\) Solve for \(x\): \(x \approx 1.05 \times 10^{-5} \mathrm{M}\) Now, calculate the pH using: \(\mathrm{pH = -\log[H^{+}]}\) \(\mathrm{pH} = -\log(1.05 \times 10^{-5}) \approx 4.98\) #b) The pH after adding 0.1 mole of HCl per liter is approximately 4.98#

Key concepts

Acid-Base Equilibrium

Understanding acid-base equilibrium is key to solving pH-related problems. At the heart of this is how acids and bases interact in water. When an acid, such as acetic acid (\(\mathrm{CH}_3\mathrm{COOH}\)), dissolves in water, it partially dissociates into its conjugate base (\(\mathrm{CH}_3\mathrm{COO}^-\)) and releases hydrogen ions (\(\mathrm{H}_3\mathrm{O}^+\)).The balance between reaction products and reactants determines equilibrium. This dynamic balance is captured by the acid dissociation constant (\(\mathrm{K_a}\)). For acetic acid, this equilibrium reaction is:\[\mathrm{CH}_3\mathrm{COOH} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}_3\mathrm{O}^+\]The equilibrium constant provides insight into the extent of dissociation. A small \(\mathrm{K_a}\) like \(1.75 \times 10^{-5}\) indicates weak dissociation, typical of weak acids like acetic acid.

ICE Table

The ICE table is a powerful tool for visualizing concentration changes in equilibrium reactions. ICE stands for Initial, Change, and Equilibrium — each representing stages of the reaction.1. **Initial**: Set up the initial concentrations before the reaction shifts. For a solution of acetic acid \(0.5 \mathrm{M}\) and acetate ion \(0.5 \mathrm{M}\), the hydronium concentration is typically \(0\).2. **Change**: Denote changes in concentrations as \(x\). Weak acids like acetic acid barely change, ensuring easier calculations.3. **Equilibrium**: Sum initial and change for equilibrium concentrations. Use these in the \(\mathrm{K_a}\) formula:\[\mathrm{K_a} = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} = \frac{x(0.5 + x)}{0.5 - x}\]In most cases with weak acids, \(x\) is a minor adjustment, simplifying calculations.

Acetic Acid

Acetic acid is a common weak acid with the formula \(\mathrm{CH}_3\mathrm{COOH}\). It is the main component of vinegar and widely used in lab solutions.Due to its weak acidic nature, acetic acid partially dissociates in water. This incomplete dissociation means it reaches equilibrium without converting all molecules into hydrogen ions and acetate ions.The dissociation level is governed by the acid dissociation constant (\(\mathrm{K_a}\)), a unique value for each weak acid. For acetic acid, \(\mathrm{K_a}\) is \(1.75 \times 10^{-5}\). This small value reflects its weak acidic strength and predicts its behavior in solutions — it dissociates slightly, making it an ideal acid for studying equilibrium reactions and buffer solutions.

Buffer Solution

A buffer solution contains a weak acid like acetic acid and its conjugate base, in this case, acetate ion (\(\mathrm{CH}_3\mathrm{COO}^-\)). Buffer solutions resist drastic pH changes upon addition of small amounts of strong acids or bases.Here, the concentrations of the acid and conjugate base regulate \([\mathrm{H}^+]\) changes. When \(\mathrm{HCl}\) is added, the acetate ion reacts with hydrogen ions to form more acetic acid. This consumption maintains near constant \([\mathrm{H}^+]\), mildly altering pH.To calculate the impact of added acid on such a buffer, adjust initial concentrations with the added component:- Acetic Acid increases: \(0.5 + 0.1 = 0.6 \mathrm{M}\)- Acetate decreases: \(0.5 - 0.1 = 0.4 \mathrm{M}\)This highlights the buffer's capacity for neutralizing added acids and explains its importance in chemical solutions, maintaining stable environments.