Problem 47 - Advanced Problems in Organic Chemistry for Competitive Examinations

Question

What will be the \(\mathrm{pH}\) of an acetate-acetic acid solution when the ratio of \(\left[\mathrm{CH}_{3} \mathrm{CO}_{2}\right] /$$\left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right]\) is \(10 ?\) (A table of \(\mathrm{pK}\) data is given below.) $$ \begin{array}{|lc|} \hline \text { Some useful } & \mathrm{pK}_{\mathrm{a}} \text { values } \\ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} & 4.76 \\ \mathrm{H}_{3} \mathrm{PO}_{4} & 2.2 \\ \mathrm{H}_{2} \mathrm{PO}^{\rho} & 7.2 \\ \mathrm{HPO}_{4}^{2} & 12.4 \\ \hline \end{array} $$ (a) \(5.76\) (b) \(4.76\) (c) \(3.76\) (d) \(1.76\)

Step-by-Step Solution

Answer

The pH of the solution is 5.76, option (a).

Step 1: Identify the Known Values

We know from the problem that the ratio \( \left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right] / \left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right] \) is 10 and the \( \mathrm{pK}_{\mathrm{a}} \) of acetic acid is 4.76.

Step 2: Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \[ \mathrm{pH} = \mathrm{pK}_a + \log \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right) \]Where \([\mathrm{A}^-]\) is the concentration of the acetate ion, and \([\mathrm{HA}]\) is the concentration of acetic acid.

Step 3: Substitute Values into the Equation

Substitute \( \mathrm{pK}_{\mathrm{a}} = 4.76 \) and the ratio \( \left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right] / \left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right] = 10 \) into the equation:\[ \mathrm{pH} = 4.76 + \log(10) \]

Step 4: Calculate the Logarithm

Calculate \( \log(10) \). Since the logarithm of 10 in base 10 is 1, we have:\[ \log(10) = 1 \]

Step 5: Find the Final pH

Substitute \( \log(10) = 1 \) into the equation:\[ \mathrm{pH} = 4.76 + 1 = 5.76 \].

Key Concepts

pH calculation

Calculating the pH of a solution involves determining how acidic or basic that solution is. The pH scale ranges from 0 to 14, where lower values indicate acidic solutions, higher values indicate basic solutions, and a pH of 7 is considered neutral. In the context of an acetate-acetic acid solution, we use the Henderson-Hasselbalch equation, a convenient way to calculate the pH of a buffer solution. This equation is particularly useful when dealing with solutions that involve a weak acid and its conjugate base.
For the acetate-acetic acid solution, we need two key pieces of information:

• The acid dissociation constant, expressed as pKa, which indicates the strength of the weak acid.
• The ratio of the concentrations of the base form (acetate ion, \([\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}]\)) and acid form (acetic acid, \([\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}]\)).

By entering these into the Henderson-Hasselbalch equation, \( \text{pH} = \text{pK}_a + \log \left( \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\right)\), we determine how close the pH is to the pKa value, thus gauging the solution's acidity or basicity. In our exercise, given that \( \text{pK}_a\) is 4.76 and the ratio is 10, the solution turns out to be more basic due to the higher concentration of the conjugate base.

acid-base equilibrium

An acid-base equilibrium involves the balance between acids and bases in a solution. This concept is crucial in understanding how buffers work, such as the acetate-acetic acid buffer. In equilibrium, the forward reaction (where acid disassociates to form base and hydrogen ions) occurs at the same rate as the reverse reaction (where base and hydrogen ions recombine to form the acid).
The ability of a buffer solution to resist changes in pH upon the addition of small amounts of acid or base is its defining characteristic.

• In our exercise, acetic acid (\[CH_3CO_2H\]) acts as a weak acid, partially dissociating in water to form hydrogen ions and the acetate ion (\[CH_3CO_2^-\]).
• This dissociation reaches equilibrium, where the rate of dissociation of acetic acid equals the rate of recombination of acetate ions and hydrogen ions.

Using the ratio of \( \left[\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\right] / \left[\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\right] = 10\), we can visualize that the base is present in significantly larger amounts, skewing the equilibrium towards a more basic state than the neutral midpoint of around pKa.

chemical equilibrium

Chemical equilibrium refers to the state in which both the reactants and products of a chemical reaction are present in concentrations that have no further tendency to change with time. It represents a balance between the forward and reverse reactions. In the context of our problem, chemical equilibrium is reached when the rate of formation of acetate ions (\[\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}\]) balances with the rate of reformation of acetic acid (\[\mathrm{CH}_{3}\mathrm{CO}_2\mathrm{H}\]).
This equilibrium is described by the expression of the equilibrium constant, Ka, which is specific for a given temperature and is defined as the ratio of the concentration of products to reactants during equilibrium:

• For our system, this translates into an equation involving the concentration of acetate ions, acetic acid, and hydrogen ions.
• When we apply the Henderson-Hasselbalch equation, we account for this equilibrium by considering the pKa, which is the \(-\log(K_a)\), reflecting the acid's strength.

By understanding this dynamic equilibrium, we appreciate how the solution's pH can be precisely controlled even as reactions proceed both ways under constant conditions. This knowledge of chemical equilibrium is fundamental to predicting the behavior of buffer solutions and their capacity to stabilize pH levels effectively.