Problem 57 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The solubility of a salt of weak acid \((A B)\) at \(\mathrm{pH} 3\) is \(\boldsymbol{Y} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}\). The value of \(\boldsymbol{Y}\) is (Given that the value of solubility product of \(\boldsymbol{A B}\) \(\left(K_{s p}\right)=2 \times 10^{-10}\) and the value of ionization constant of \(\left.\boldsymbol{H B}\left(K_{a}\right)=1 \times 10^{-8}\right)\)

Step-by-Step Solution

Answer

The value of \(Y\) is 141.

Step 1: Understand the Solubility Equilibrium

The salt \(AB\) dissociates completely in solution as \(A^+\) and \(B^-\). The weak acid version \(HB\) partially ionizes in the water. This means we have an equilibrium:\[AB \rightleftharpoons A^+ + B^-\]

Step 2: Write the Expression for Ksp

For the dissociation equilibrium of the salt \(AB\), the solubility product \(K_{sp}\) can be written as:\[K_{sp} = [A^+][B^-] = (s)(s) = s^2\]where \(s\) is the solubility of the salt in \(\text{mol/L}\). Here, \(K_{sp} = 2 \times 10^{-10}\).

Step 3: Write the Expression for Ka

For the ionization of the weak acid \(HB\), the ionization constant \(K_a\) is given by:\[K_{a} = \frac{[H^+][B^-]}{[HB]}\]We assume \([B^-] = s\) due to dissolving of \(AB\), and \([H^+] = 10^{-3}\ \text{mol/L}\) because the pH is 3.

Step 4: Calculate [B^-] in Equilibrium with the Acid

Since \(HB\) partially ionizes, we include the contribution of \([B^-]\) from this ionization. We set up the relation:\[K_a = \frac{(10^{-3})(s)}{(s)} = 1 \times 10^{-8}\]Simplifying gives \(10^{-3} = s \times 10^{-8}\) or \(s = \frac{10^{-3}}{10^{-8}} = 10^5\). This assumes the contribution from \(HB\) is negligible in comparison to \(s\).

Step 5: Calculate the Total Solubility

Now, calculate the total solubility \([B^-]\) considering both sources at equilibrium:\[[B^-] = s = Y \times 10^{-3}\]From Step 2, \(K_{sp} = s^2 = 2 \times 10^{-10}\), thus:\(s = \sqrt{2 \times 10^{-10}} = \sqrt{2} \times 10^{-5}\ \text{mol/L}\)Convert this to match the format \(Y \times 10^{-3}\):\[Y \times 10^{-3} = \sqrt{2} \times 10^{-5}\]Thus, \(Y = \sqrt{2} \times 10^2\).

Step 6: Calculate Y

Compute \(\sqrt{2} \approx 1.414\), then:\[Y = 1.414 \times 10^2 = 141.4\]Round to the nearest whole number gives \(Y = 141.\)

Key Concepts

Solubility Product

In the context of chemistry, particularly when working with salts, the solubility product ( \( K_{sp} \) ) plays a crucial role. The solubility product is a constant that denotes the extent to which a compound can dissolve in water, indicating how much of the salt disassociates into its ions in solution. For any salt like \( AB \) that dissociates into \( A^+ \) and \( B^- \) , the solubility product is expressed in terms of the concentrations of these ions: \[ K_{sp} = [A^+][B^-] = s^2 \] Here, \( s \) represents the solubility of the salt in mol/L. The larger the value of \( K_{sp} \) , the more soluble the compound is under standard conditions. However, in this problem, since \( K_{sp} = 2 \times 10^{-10} \) , it indicates a low solubility. It is important to represent the equilibrium between dissolved ions accurately as it dictates how the substance behaves in aqueous solutions.

Ionization Constant

The ionization constant, often denoted as \( K_a \) , is associated with weak acids and represents the equilibrium constant for the dissociation of the acid into its ions. A weak acid, unlike strong acids, does not completely dissociate in water. Instead, it sets up an equilibrium similar to this: \[ HB \rightleftharpoons H^+ + B^- \] The expression for the ionization constant \( K_a \) for weak acid \( HB \) is given by: \[ K_{a} = \frac{[H^+][B^-]}{[HB]} \] Given the problem, we know \( K_{a} = 1 \times 10^{-8} \) and that the acid only partially ionizes, which is indicative of its weak nature. The concentration of \( [H^+] = 10^{-3} \text{ mol/L} \) given by the pH value, emphasizes how these conditions affect the extent of ionization.

Weak Acid

A weak acid is a type of acid that does not fully ionize in water. This incomplete ionization results in an equilibrium between the undissociated acid molecules and the ions produced. In practical terms, this means that only some of the acid turns into \( H^+ \) and its conjugate base in water: \[ HB \rightleftharpoons H^+ + B^- \] For weak acids, the ionization constant \( K_a \) helps to gauge the strength of the acid. A smaller \( K_a \) suggests weaker ionization, meaning less of the acid dissociates into ions.

• The behavior of weak acids is crucial for calculating total solubility, particularly when the acid's dissociation overlaps with a salt's solubility.
• In this exercise, understanding the weak acid's dynamics assists in determining how much \( B^- \) ions stem from each source — both the salt and the acid.
• By evaluating the weak acid's contribution, we ensure that we do not overestimate ion concentrations in solution.

This check is integral for accurate calculations and understanding the interplay between solubility and ionization.