Problem 60 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

What is the \(\mathrm{pH}\) of a \(0.50 \mathrm{M}\) aqueous \(\mathrm{NaCN}\) solution? \(\mathrm{p} K_{b}\) of \(\mathrm{CN}^{-}\)is \(4.70\)

Step-by-Step Solution

Answer

The pH of the solution is 11.03.

Step 1: Identify the Reaction

Sodium cyanide (NaCN) dissociates in water to form sodium ions (Na⁺) and cyanide ions (CN⁻). The cyanide ion in water acts as a base and undergoes hydrolysis: \[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \]

Step 2: Calculate the pOH

Use the given \(\mathrm{p} K_{b}\) of CN⁻ to find the concentration of OH⁻. The equation for the base dissociation constant is \[ K_b = 10^{-pK_b} = 10^{-4.70} \]Set up the equation for the equilibrium expression: \[ K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]} \]Assume initial concentration of \([\text{CN}^-] = 0.50 \text{ M}\) and let \(x\) be the change in concentration for \([\text{OH}^-]\): \[ K_b = \frac{x^2}{0.50 - x} \]Since \(K_b\) is small, \(0.50 - x \approx 0.50\): \[ x^2 = K_b \cdot 0.50 \]Substitute \(K_b\): \[ x^2 = 10^{-4.70} \cdot 0.50 \] \[ x = \sqrt{0.50 \cdot 10^{-4.70}} \]

Step 3: Calculate the OH⁻ Concentration

Solve for \(x\) to find \([\text{OH}^-]\): Perform the calculation: \[ x = \sqrt{0.50 \cdot 10^{-4.70}} \approx 1.08 \times 10^{-3} \text{ M}\]Thus, \([\text{OH}^-] = 1.08 \times 10^{-3} \text{ M}\).

Step 4: Find the pOH Value

Calculate the pOH from the \([\text{OH}^-]\) concentration:\[ \text{pOH} = -\log_{10}[\text{OH}^-] \] Substitute the concentration of OH⁻:\[ \text{pOH} = -\log_{10}(1.08 \times 10^{-3}) \approx 2.97 \]

Step 5: Calculate the pH

Use the relationship between pH, pOH, and 14:\[ \text{pH} + \text{pOH} = 14 \] Substitute the calculated pOH:\[ \text{pH} = 14 - 2.97 = 11.03 \]

Step 6: Final Answer

The \(\text{pH}\) of the solution is 11.03.

Key Concepts

Base Dissociation Constant

The base dissociation constant, often represented as \(K_b\), is a crucial value when analyzing how a base ionizes in water to produce hydroxide ions \(\text{OH}^-\). It helps to determine the strength of a base in solution. In the context of cyanide ions \((\text{CN}^-)\), which arise from sodium cyanide (NaCN) dissolving in water, the base dissociation constant can be calculated using the formula:\[ K_b = 10^{-\text{p}K_b} \]The given pK\(_b\) for cyanide ions is 4.70. Therefore, you can find \(K_b\) by plugging 4.70 into the equation:\[ K_b = 10^{-4.70} \]As \(K_b\) is a relatively small number, it indicates that cyanide ions are a weak base. This means they do not dissociate completely in water. Instead, equilibrium is established between the reactants and products. Understanding the base dissociation constant aids in calculating other related properties, such as the concentration of hydroxide ions \([\text{OH}^-]\), which is key to determining the pH of solutions related to weak bases like cyanide.

Cyanide Ion

Cyanide ions \((\text{CN}^-)\) are negatively charged ions that result from the dissociation of sodium cyanide \((\text{NaCN})\) in water. When \(\text{NaCN}\) dissolves, it releases sodium ions \((\text{Na}^+)\) and cyanide ions \((\text{CN}^-)\), specializing in creating a basic solution. This is because the cyanide ion acts as a Brønsted-Lowry base, implying that it tends to accept protons \((\text{H}^+)\) from water, leaving behind hydroxide ions \((\text{OH}^-).\)The dissociation and hydrolysis reaction of \(\text{CN}^-\) in water can be represented by the equation:\[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \]This equilibrium reaction shows how the cyanide ion picks up a proton from water to form hydrogen cyanide \((\text{HCN})\), producing more \(\text{OH}^-\) and thus making the solution basic. This process is critical for calculating the concentrations required for pH assessments in weak base solutions like the ones involving \(\text{CN}^-\).

Hydrolysis

Hydrolysis refers to the chemical breakdown of a compound due to the reaction with water and plays a pivotal role when cyanide ions \((\text{CN}^-)\) are present in the solution. In the context of sodium cyanide \((\text{NaCN})\), hydrolysis occurs when cyanide ions interact with water, accepting protons and forming hydroxide ions \((\text{OH}^-\)).The hydrolysis reaction is the key:\[ \text{CN}^- + \text{H}_2\text{O} \rightleftharpoons \text{HCN} + \text{OH}^- \]Here, the cyanide ion undergoes a reaction that creates a weak acid, hydrogen cyanide \((\text{HCN})\), and more \(\text{OH}^-\) ions. This increase in \(\text{OH}^-\) ions is why the solution becomes basic, elevating its pH. Hydrolysis reactions like this are essential when finding the pH because they determine how much the solution leans towards being basic or acidic. Understanding this concept helps demystify why solutions formed from certain salts, particularly those containing weak bases like \(\text{CN}^-\), result in higher pH values.