Problem 20 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

$$ \begin{aligned} &\mathrm{NaOH} \text { is a strong base. What will be } \mathrm{pH} \text { of } 5.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH}\\\ &\text { solution ? }(\log 2=0.3) \end{aligned} $$ (a) \(14.00\) (b) \(13.70\) (c) \(13.00\) (d) \(12.70\)

Step-by-Step Solution

Answer

The pH of the solution is closest to 12.70, option (d).

Step 1: Identify the concentration of OH- ions

Since NaOH is a strong base, it dissociates completely in water. The concentration of NaOH is given as \(5.0 \times 10^{-2} \ M\). Thus, the concentration of \( \text{OH}^- \) ions is also \(5.0 \times 10^{-2} \ M\).

Step 2: Calculate the pOH of the solution

The pOH can be calculated using the formula \( \text{pOH} = -\log [\text{OH}^-] \). Given \([\text{OH}^-] = 5.0 \times 10^{-2} \), we find the log as follows: \[ \log (5.0 \times 10^{-2}) = \log (5) + \log (10^{-2}) = \log(5) - 2\]. Since \( \log (5) = \log (5 \times 10^0) = 0.3 + \log(10^0) = 0.3 \), the expression becomes \( \log (5.0 \times 10^{-2}) = 0.3 - 2 = -1.7\). Thus, \( \text{pOH} = -(-1.7) = 1.7 \).

Step 3: Calculate the pH of the solution

The pH and pOH are related by the equation \( \text{pH} + \text{pOH} = 14 \). Use this relationship to find \( \text{pH} \): \[ \text{pH} = 14 - \text{pOH} = 14 - 1.7 = 12.3\].

Step 4: Match pH value to given choices

Review the calculated pH value of 12.3 and see which given option is closest. The closest option to 12.3 is (d) \(12.70\).

Key Concepts

Understanding Strong Bases

When we talk about strong bases, we refer to substances that completely dissociate into their ions when dissolved in water. This means they release a significant number of hydroxide (\(\text{OH}^-\)) ions, which are key players in affecting the basicity of a solution. An understanding of strong bases is essential because these substances dramatically alter the pH of a solution due to their full dissociation.

• Complete Dissociation: Strong bases dissociate 100% in aqueous solutions, meaning none of the base remains in its original form.
• High \(\text{OH}^-\) Concentration: They release a large number of hydroxide ions, making the solution highly basic.
• Examples: Common strong bases include \( ext{NaOH}\) (sodium hydroxide) and \( ext{KOH}\) (potassium hydroxide).

Sodium hydroxide (\( ext{NaOH}\)) is a classic example. As a strong base, it plays a significant role in numerous chemical processes, industrial applications, and laboratory experiments.

Sodium Hydroxide (NaOH) in Solution

Sodium hydroxide, often called caustic soda, is a widely used chemical due to its strong alkaline properties. When NaOH is dissolved in water, it separates into sodium (\( ext{Na}^+\)) and hydroxide (\(\text{OH}^-\)) ions. This dissociation is complete, meaning the concentration of hydroxide ions in the solution matches the initial concentration of the sodium hydroxide.
For this particular problem, we're given an initial concentration of \(5.0 \times 10^{-2} \, M\) for \( ext{NaOH}\). Since it dissociates completely, the concentration of the hydroxide ions is also \(5.0 \times 10^{-2} \, M\). This knowledge prepares us for the next step in solving for the pH of the solution.

• Sodium ions (\( ext{Na}^+\)) are inert concerning pH changes; they don't affect acidity or basicity.
• The significant player is the hydroxide ion, which heavily influences the solution's pH.

Understanding the role of \( ext{NaOH}\) helps us quantify the pH accurately, which is crucial for chemical calculations and reactions.

Calculating pOH and pH

In a solution containing a strong base like \( ext{NaOH}\), determining the pOH is a crucial step in calculating the pH. The pOH is a measure of hydroxide ion concentration, similar but opposite to pH which measures hydrogen ion concentration.We can calculate pOH using the formula:\[ \text{pOH} = -\log [\text{OH}^-]\]For this problem, the hydroxide ion concentration is \(5.0 \times 10^{-2} \, M\). By computing \(\log(5)\) and knowing \(\log(10^{-2}) = -2\), we arrive at \(\text{pOH} = 1.7\).
This indicates a very basic solution. Knowing that pH and pOH are related by the equation:\[ \text{pH} + \text{pOH} = 14 \]We find pH as:\[ \text{pH} = 14 - \text{pOH} = 12.3\]This calculated pH value is close to the provided option of \(12.70\), highlighting the approximations made in logarithmic calculations. This process underscores the importance of understanding both pOH and pH calculations for accurate chemical analysis.