Problem 50 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(0.1 \mathrm{M} \mathrm{NaOH}\) is titrated with \(0.1 \mathrm{M} \mathrm{H} A\) till the end point; \(K_{a}\) for \(\mathrm{H} A\) is \(5.6 \times 10^{-6}\) and degree of hydrolysis is less compared to 1. Calculate \(\mathrm{pH}\) of the resulting solution at the end point.

Step-by-Step Solution

Answer

The pH of the solution is approximately 8.63.

Step 1: Understanding the Problem

In this problem, NaOH, a strong base, is titrated with HA, a weak acid, until the endpoint. At the endpoint, the moles of the strong base are equal to the moles of the weak acid, forming the conjugate base (A-) of the weak acid (HA). The task is to calculate the pH of the solution at this point.

Step 2: Determine Reaction at Endpoint

The reaction is \[\text{HA + NaOH} \rightarrow \text{NaA + H}_2\text{O}.\]At the endpoint, all of HA is converted into its conjugate base, A-. The solution majorly contains the conjugate base (A-) of HA, which will impact the pH.

Step 3: Calculate the Concentration of A-

Since the moles of HA and NaOH are equal and that both are 0.1 M, after neutralization, the concentration of NaA (or A-) will also be 0.1 M as it fully dissociates in water.

Step 4: Determine OH- Concentration via Hydrolysis of A-

A- comes from the weak acid HA, thus it will undergo hydrolysis: \[\text{A}^- + \text{H}_2\text{O} \leftrightarrow \text{HA} + \text{OH}^-\]The hydrolysis constant, \(K_h\), can be found using \[K_w = K_a \cdot K_b\] or \[K_h = \frac{K_w}{K_a},\] where \(K_w\) is the ion-product of water, \(1.0 \times 10^{-14} \text{ mol}^2/\text{L}^2\).

Step 5: Calculate K_h and Use It to Find [OH-]

Calculate \(K_h\):\[K_h = \frac{1.0 \times 10^{-14}}{5.6 \times 10^{-6}} = 1.7857 \times 10^{-9}\]Assuming the degree of hydrolysis is small, use the expression for \(K_h\):\[K_h = [OH^-]^2/C_{A^-},\] simplifying gives:\[[OH^-] = \sqrt{K_h \cdot C_{A^-}}\]Substitute \(K_h\) and \(C_{A^-} = 0.1 \text{ M}\): \[[OH^-] = \sqrt{1.7857 \times 10^{-9} \times 0.1} = 4.228 \times 10^{-6} \text{ M}\]

Step 6: Calculate pH from [OH-]

From [OH-], calculate pOH:\[\text{pOH} = -\log_{10} [OH^-] = -\log_{10} (4.228 \times 10^{-6}) = 5.374\]Convert pOH to pH:\[\text{pH} = 14 - \text{pOH} = 14 - 5.374 = 8.626\]

Step 7: Conclusion

The pH of the solution at the endpoint when 0.1 M NaOH is titrated with 0.1 M HA is calculated to be approximately 8.63.

Key Concepts

pH calculation

The calculation of pH plays a crucial role in understanding the acidity or basicity of a solution. In the context of acid-base titration, particularly at the endpoint, calculating the pH helps determine the characteristics of the resulting solution. In this particular exercise, we have a scenario where 0.1 M Sodium Hydroxide (NaOH), a strong base, is neutralized by 0.1 M HA, a weak acid. At the endpoint, equal moles of NaOH and HA react, converting into their conjugate base form, A-. When we focus on the solution at this point, the concentration of hydroxide ions (OH^-) becomes a primary concern as it dictates the pH of the solution.

Here's how it is done:

• Determining hydroxide ion concentration ([OH^-]) via hydrolysis of the conjugate base (A-),
• Calculating the pOH from [OH^-], and
• Converting pOH to pH using the formula: \[ \text{pH} = 14 - \text{pOH} \]

Each step embodies a transition from one concentration measurement to another, ultimately revealing the nature of the solution as either acidic, neutral, or basic.

weak acid

Weak acids, such as HA in this exercise, do not fully dissociate in solution. This means that when HA is dissolved in water, only a fraction of its molecules release a hydrogen ion (H+) to form the hydronium ion ( H_3O^+ ).

The degree of dissociation is an essential consideration:

• The presence of these hydrogen ions decreases the solution's pH,
• Its equilibrium state can be simply represented via an equilibrium constant ( K_a ).

In our example, HA has the dissociation constant K_a = 5.6 imes 10^{-6} , indicating a relatively low level of ionization in water. This nature significantly impacts not only the pH of its solutions but also plays a critical role in its behavior during reactions such as titrations when transformed into its conjugate base.

conjugate base

When weak acids like HA are involved in a reaction such as titration, they convert into their conjugate base (A-) once they donate their hydrogen ion (H+) to reactants like strong bases. During a titration involving a strong base and a weak acid, the creation of the conjugate base is crucial.

• At the endpoint of titration, all the weak acid HA has converted into its conjugate base, A-.
• A- will not just "sit" in the solution.
• It hydrolyzes, reacting with water to form OH-, influencing the final pH of the solution.

This transformation from a weak acid to a corresponding conjugate base at the equivalence point elucidates why even after neutralization, the solution maintains certain basic characteristics, as calculated in this exercise with the pH value of approximately 8.63.

hydrolysis constant

The hydrolysis constant (K_h) is key in understanding the behavior of a conjugate base derived from a weak acid during its interaction with water. In the equation HA+ NaOH \rightarrow NaA + H_2O, A- undergoes hydrolysis, impacting the solution's hydroxide ion concentration and hence its pH.

To solve this, one uses:

• The relationship between K_w (ion-product constant of water) and K_a: \[ K_h = \frac{K_w}{K_a} \]
• Allowing us to compute K_h = 1.7857 \times 10^{-9} for A- in this exercise.

Once K_h is known, it aids in deriving the hydroxide ion concentration [OH^-], by assuming the degree of hydrolysis is negligible. This simplifies calculations, assisting in accurately determining the endpoint pH using straightforward relationships between the constants and concentrations involved.