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

The pH of the NaCN solution is approximately 11.5.

Step 1: Calculate Kb

Recall that \(\mathrm{pK}_b = -\log_{10}(K_b)\). Given \(\mathrm{pK}_b = 4.7\), we can calculate \(K_b\) by rearranging the formula: \(K_b = 10^{-4.7}\). Using a calculator, find \(K_b \approx 2.0 \times 10^{-5}\).

Step 2: Use ICE Table for CN- Ion

Write the equilibrium reaction for the ionization of \(\mathrm{CN}^-\), which is \(\mathrm{CN}^- + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCN} + \mathrm{OH}^-\). Using the ICE (Initial, Change, Equilibrium) Table, we have the initial concentration of \(\mathrm{CN}^- = 0.5 \, \mathrm{M}\), \(\mathrm{HCN} = 0\), and \(\mathrm{OH}^- = 0\). At equilibrium let the change be \(x\), so the equilibrium concentrations are \(\mathrm{HCN} = x\), and \(\mathrm{OH}^- = x\).

Step 3: Set Up the Equilibrium Expression

The equilibrium expression for the reaction is \(K_b = \frac{[\mathrm{HCN}][\mathrm{OH}^-]}{[\mathrm{CN}^-]}\). Substitute the equilibrium concentrations: \(2.0 \times 10^{-5} = \frac{x^2}{0.5 - x}\). Assuming \(x\) is much smaller than 0.5, simplify to \(2.0 \times 10^{-5} = \frac{x^2}{0.5}\) which gives \(x^2 = 1.0 \times 10^{-5}\).

Step 4: Solve for x

Solve the equation \(x^2 = 1.0 \times 10^{-5}\) to find \(x\). Taking the square root, \(x = \sqrt{1.0 \times 10^{-5}} = 3.16 \times 10^{-3}\). So \(x = [\mathrm{OH}^-] = 3.16 \times 10^{-3} \, \mathrm{M}\).

Step 5: Calculate pOH

To find the \mathrm{pOH}\, use the formula \(\mathrm{pOH} = -\log_{10}([\mathrm{OH}^-])\). Substituting the value of \(x = 3.16 \times 10^{-3}\) gives \(\mathrm{pOH} \approx 2.5\).

Step 6: Determine pH

Using the relationship \(\mathrm{pH} + \mathrm{pOH} = 14\), solve for \mathrm{pH}\: \(\mathrm{pH} = 14 - \mathrm{pOH}\). Substituting \(\mathrm{pOH} \approx 2.5\), we get \(\mathrm{pH} = 14 - 2.5 = 11.5\).

Key Concepts

Kb calculation

When dealing with weak bases like cyanide ion (\(\text{CN}^-\)), calculating its base dissociation constant (\(K_b\)) is crucial for understanding its strength in water. The \(K_b\) value indicates how well a base can accept protons in solution. To find \(K_b\), we use the given \(\text{p}K_b\) value. The formula that links them together is:

• \(\text{p}K_b = -\log_{10}K_b\)

From this, we can deduce:

• \(K_b = 10^{-\text{p}K_b}\)

In our scenario, \(\text{p}K_b\) for \(\text{CN}^-\) is 4.7, so:

• \(K_b = 10^{-4.7} \approx 2.0 \times 10^{-5}\)

Having \(K_b\) helps us set up equations describing how the base will behave in water.

ICE table

The ICE table is a wonderful tool to visualize how the concentrations of substances change as a chemical reaction approaches equilibrium. Here's how the ICE table unfolds:

• **I**nitial: Start with the concentrations before the reaction takes place. For \(\text{CN}^-\), it's 0.5 M, whereas initially \(\text{HCN}\) and \(\text{OH}^-\) are both 0.
• **C**hange: As the reaction proceeds, \(\text{CN}^-\) decreases by \(x\) and both \(\text{HCN}\) and \(\text{OH}^-\) increase by \(x\).
• **E**quilibrium: It gives us the concentrations at equilibrium. From the table, \([\text{HCN}] = x\) and \([\text{OH}^-] = x\).

The ICE table simplifies tracking changes, helping us understand how the equilibrium state is achieved in the reaction.

equilibrium expression

This expression is key to tying together the concentrations of the reactants and products once equilibrium is reached. It's governed by the formula:

• \(K_b = \frac{[\text{HCN}][\text{OH}^-]}{[\text{CN}^-]}\)

Substituting the equilibrium concentrations yields:

• \(2.0 \times 10^{-5} = \frac{x^2}{0.5 - x}\)

Due to \(x\) being typically small compared to the initial concentration, the expression can be further simplified by assuming \(0.5 - x \approx 0.5\). Thus:

• \(2.0 \times 10^{-5} = \frac{x^2}{0.5}\)

Resolving this gives \(x^2 = 1.0 \times 10^{-5}\), letting us find \(x\). This step ensures accurate tracking of equilibrium in the system, essential for subsequent calculations.

pOH to pH conversion

Converting \(\text{pOH}\) to \(\text{pH}\) is significant as it completes the understanding of the solution's acidity or basicity. The interconnection is very close, made simple with the following relationship:

• \(\text{pH} + \text{pOH} = 14\)

Once \(\text{OH}^-\) concentration is known, \(\text{pOH}\) is found using:

• \(\text{pOH} = -\log_{10}([\text{OH}^-])\)

Substituting \([\text{OH}^-] = 3.16 \times 10^{-3} \space \text{M}\), we calculate \(\text{pOH} \approx 2.5\). Using the conversion formula, we determine:

• \(\text{pH} = 14 - 2.5 = 11.5\)

This conclusive step gives us the solution's \(\text{pH}\), verifying it as a basic solution and resolving the exercise effectively.