Problem 174 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(\mathrm{pK}_{\mathrm{b}}\) for \(\mathrm{CN}-\) at \(25^{\circ} \mathrm{C}\) is \(4.7\), the \(\mathrm{pH}\) of \(0.5 \mathrm{M}\) aqueous NaCN solution is (a) 10 (b) \(11.5\) (c) 11 (d) 12

Step-by-Step Solution

Answer

(b) 11.5

Step 1: Understand the relationship between pKb and pKa

The relationship between the base dissociation constant (pK_b) and the acid dissociation constant (pK_a) of a conjugate acid-base pair is given by the equation: \( \mathrm{pK_a} + \mathrm{pK_b} = 14 \). Given the pK_b for \( \mathrm{CN}^- \) is 4.7, we can find pK_a for its conjugate acid \( \mathrm{HCN} \).

Step 2: Calculate pKa using pKb

Using the relation \( \mathrm{pK_a} + \mathrm{pK_b} = 14 \), we can calculate pK_a:\[ \mathrm{pK_a} = 14 - \mathrm{pK_b} = 14 - 4.7 = 9.3 \]

Step 3: Set up the equation for ionization of NaCN

NaCN is a salt of a weak acid (HCN) and a strong base (NaOH). In solution, CN\(^-\) will slightly hydrolyze to form OH\(^-\) ions, which is the reverse of the dissociation of HCN. We use the expression for equilibrium constant to find the concentration of OH\(^-\).

Step 4: Use pKa to find the equilibrium expression

Let \([\mathrm{OH}^-]\) at equilibrium be \(x\). The hydrolysis reaction is \( \mathrm{CN}^- + \mathrm{H_2O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^- \). The expression is: \( K_w / K_a = x^2 / (0.5 - x) \).

Step 5: Simplify and solve for x

Since \(K_w = 10^{-14}\) and \(K_a = 10^{-9.3}\), \(K_w / K_a = 10^{-4.7}\). Assume that \(x\) is small compared to 0.5, we have: \[ x^2 \approx 0.5 \times 10^{-4.7} \rightarrow x^2 = 0.5 \times 2 \times 10^{-5} = 10^{-5} \] Thus, \( x = 10^{-2.5} = 3.2 \times 10^{-3} \) mol/L.

Step 6: Calculate pOH and pH

Calculate \(\mathrm{pOH} = -\log_{10}([\mathrm{OH}^-]) = -\log_{10}(3.2 \times 10^{-3}) \approx 2.5\). Finally, calculate \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.5 = 11.5\).

Step 7: Verify the answer

The closest given answer to the calculated pH of 11.5 is option (b) 11.5.

Key Concepts

pK_b and pK_a relationship

The relationship between the base dissociation constant, known as pK_b, and its counterpart for acids, pK_a, is instrumental in understanding chemical equilibria for solutions involving acidic and basic components. The formula connecting these constants is crucial:

• \( \mathrm{pK_a} + \mathrm{pK_b} = 14 \).

This expression originates from the auto-ionization of water, where the ion product \( K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14} \). Knowing the pK_b of a base allows us to calculate the pK_a of its conjugate acid, which is vital in predicting the behavior of salts in solution. For example, with

• \( \mathrm{pK_b} = 4.7 \) for \( \mathrm{CN}^- \),
• \( \mathrm{pK_a} \) for \( \mathrm{HCN} \) becomes \( 14 - 4.7 = 9.3 \).

This relationship enables chemists to derive other necessary parameters to solve equilibrium problems, particularly in solutions of salts derived from weak acids and strong bases.

chemical equilibrium

Chemical equilibrium is a fundamental concept in chemistry that describes a state where both the forward and reverse reactions occur at the same rate, resulting in a stable concentration of products and reactants over time. In the context of the NaCN solution, when dissolved, the equilibrium involves the partial ionization of CN\(^-\) in water:

• \( \mathrm{CN}^- + \mathrm{H_2O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^- \).

This reversible reaction establishes an equilibrium between the reactants and products, affected by their concentrations. The presence of a strong base and weak acid influences how the equilibrium is set, directing the reaction's balance and ultimately influencing the pH of the solution. Calculating this equilibrium involves applying the equilibrium constant expressions derived from the pK_a values, which can then be used to deduce concentrations and pH.

hydrolysis of salts

The hydrolysis of salts occurs when a salt reacts with water and affects the pH of the solution. This process is pronounced in salts formed from a strong base and a weak acid, like sodium cyanide (NaCN). In our case, the CN\(^-\) ion undergoes hydrolysis, producing hydroxide ions (OH\(^-\)) and affecting the basicity of the solution:

• \( \mathrm{CN}^- + \mathrm{H_2O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^- \).

The hydroxide ions increase the solution's pH, making it basic. The extent of hydrolysis depends on the strength of the acid from which the salt is derived. Here, knowing the pK_a indicates the degree of ionization and thus the extent of hydrolysis, and ultimately the pH. The expression:

• \( K_w / K_a = \frac{x^2}{0.5 - x} \)

helps in determining the concentration of hydroxide ions, which when calculated, reveals the solution's pH.

strong base and weak acid salt solution

A solution of a salt that is derived from a strong base and a weak acid, such as sodium cyanide (NaCN), results in an alkaline or basic solution. This characteristic stems from the complete dissociation in water of the strong base component, sodium hydroxide (NaOH), and the incomplete ionization of the weak acid, hydrocyanic acid (HCN):

• The former completely dissociates into sodium (Na\(^+\)) and hydroxide ions (OH\(^-\)), fully contributing to the basicity.
• The latter, HCN, only partially reforms from its ionized state because it is a weak acid.

The pH of such a solution is higher than 7, indicating a basic nature. Calculating pH involves understanding the extent of hydrolysis of the weak acid component. With NaCN:

• The pH is greatly influenced by the equilibrium established between the \( \mathrm{CN}^- \) and the formation of OH\(^-\) ions in the solution.

These factors lead to the pH level calculated previously reaching values like 11.5, illustrating the significant impact of the strong base portions over the weak acid in the solution. This understanding is key for predicting and controlling the behavior of such solutions in various chemical contexts.