Question

The pKa pf HCN is 9.30. The pH of a solution prepared by mixing \(2.5\) moles of \(\mathrm{KCN}\) and \(2.5 \mathrm{moles}\) of \(\mathrm{HCN}\) in water and making up the total volume to \(500 \mathrm{ml}\) is: (a) \(9.30\) (b) \(8.30\) (c) \(7.30\) (d) \(10.30\)

Short answer

The pH of the solution is 9.30, option (a).

Step-by-step solution

Understanding the System

We are given a solution that contains equal moles of HCN (a weak acid) and KCN (which supplies CN⁻, the conjugate base). This is a buffer system because both the weak acid and its conjugate base are present.

Applying the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution: \[ \text{pH} = \text{pKa} + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] where \( [\text{A}^-] \) is the concentration of the conjugate base and \( [\text{HA}] \) is the concentration of the acid. In this case, \([\text{A}^-] = [\text{CN}^-]\) and \([\text{HA}] = [\text{HCN}]\).

Calculating Concentrations

Since the total volume of the solution is 0.500 L, the concentrations of HCN and CN⁻ in the solution are: \([\text{HCN}] = \frac{2.5 \text{ moles}}{0.500 \text{ L}} = 5 \text{ M}\) and \([\text{CN}^-] = \frac{2.5 \text{ moles}}{0.500 \text{ L}} = 5 \text{ M}\). Thus, \([\text{A}^-] = [\text{HA}] \).

Calculating the pH

Substitute the values into the Henderson-Hasselbalch equation: \[ \text{pH} = 9.30 + \log \left( \frac{5}{5} \right) = 9.30 + \log(1) = 9.30 + 0 \] So, the pH of the solution is 9.30.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a remarkable tool in chemistry for calculating the pH of a buffer solution. A buffer solution is special because it resists changes in pH when small amounts of acid or base are added. This equation is especially useful when both a weak acid and its conjugate base, or a weak base and its conjugate acid, are present in a solution. The equation is expressed as: \[ \text{pH} = \text{pK}_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \] Here, \( \text{pK}_a \) is a constant value that signifies the strength of the weak acid. - \([\text{A}^-]\) represents the concentration of the conjugate base. - \([\text{HA}]\) denotes the concentration of the acid. The beauty of the Henderson-Hasselbalch equation is that it provides a straightforward way to understand how different concentrations of an acid and its conjugate base affect the pH of a solution. This is invaluable for chemists when designing solutions that need to maintain a stable pH under varying conditions.

pKa and pH relationship

The relationship between \( \text{pK}_a \) and pH is pivotal in understanding acid-base chemistry. The \( \text{pK}_a \) value is the negative logarithm of the acid dissociation constant \( K_a \), which measures the strength of an acid in solution. The lower the \( \text{pK}_a \), the stronger the acid. When using the Henderson-Hasselbalch equation, the \( \text{pK}_a \) serves as a benchmark for determining the pH of the solution when the concentrations of acid and its conjugate base are equal. - If \([\text{A}^-] = [\text{HA}]\), then the log term \( \log(1) = 0 \). Consequently, \( \text{pH} = \text{pK}_a \). - Any variation in the ratio of \([\text{A}^-]/[\text{HA}]\) results in a change from the \( \text{pK}_a \) value by the log of this ratio, thus changing the pH. Understanding \( \text{pK}_a \) provides insight into the interaction and potential equilibrium shifts between the acid and its conjugate base. This relationship is fundamental in predicting the behavior of acids and bases in different chemical environments.

Weak Acid-Base Equilibria

Equilibria involving weak acids and bases are essential concepts in chemistry for predicting the behavior of solutions. Weak acids and bases do not fully dissociate in water, meaning only a portion of the molecules break apart into ions. This partial dissociation establishes an equilibrium in the solution. Consider the weak acid hydrocyanic acid (HCN) and its conjugate base cyanide (CN⁻). When dissolved in water, HCN releases hydrogen ions \((H^+)\) and CN⁻, but not completely, leading to the following equilibrium expression: \[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \] The equilibrium constant \( K_a \) for this reaction is an indicator of the extent of dissociation. The Henderson-Hasselbalch equation can be conveniently applied to this equilibrium system. With buffered solutions, which contain both the weak acid and its conjugate base, the solution resists drastic pH changes. This quality makes buffers extremely useful in various scientific and industrial applications where maintaining a stable pH is critical. Understanding equilibria is vital for predicting how changes in concentration, temperature, or pressure can shift these equilibria, affecting the pH and behavior of the solution.