Problem 134 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following solution(s) have \(\mathrm{pH}\) between 6 and \(7 ?\) 1\. \(2 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\) 2\. \(2 \times 10^{-6} \mathrm{M} \mathrm{HCl}\) 3\. \(10^{-8} \mathrm{M} \mathrm{HCl}\) 4\. \(10^{-13} \mathrm{M} \mathrm{NaOH}\) (a) 1,2 (b) 2,3 (c) 3,4 (d) \(2,3,4\)

Step-by-Step Solution

Answer

(c) 3,4

Step 1: Understanding pH Basics

The pH is calculated using the formula \[ \mathrm{pH} = -\log[\mathrm{H}^+] \]for a solution. For acidic solutions, pH values are less than 7, while for basic solutions, they are greater than 7. A neutral solution has a pH of 7.

Step 2: Determine pH of NaOH Solutions

Calculate the pH for the NaOH solutions. Remember that NaOH is a strong base, so it completely dissociates:1. \(2 \times 10^{-6} \mathrm{M} \). Calculate the hydroxide ion concentration, and assume complete dissociation results in \([\mathrm{OH}^-] = 2 \times 10^{-6} \mathrm{M}\).2. Calculate \([\mathrm{H}^+]\) using \(K_w = [\mathrm{H}^+][\mathrm{OH}^-] = 10^{-14} \). Therefore, \([\mathrm{H}^+] = \frac{10^{-14}}{2 \times 10^{-6}} = 5 \times 10^{-9} \mathrm{M}\).3. Calculate the corresponding pH:\[ \mathrm{pH} = -\log(5 \times 10^{-9}) \approx 8.3 \]This is greater than 7, so it is excluded.4. For \( 10^{-13} \mathrm{M} \), follow a similar process: - Assume pH due to water is significant, as \([\mathrm{OH}^-] = 10^{-13} M\).- \([\mathrm{H}^+] = \frac{10^{-14}}{10^{-13}} = 10^{-1}\)- \(\mathrm{pH} = -\log(10^{-1}) = 1\) due to excess ions than pure water. However, practical pH will not differ much from water, \(\mathrm{pH} \approx 7\).The second solution is out of range; the first has its pH greater than 7.

Step 3: Determine pH of HCl Solutions

HCl is a strong acid and dissociates completely:1. For \(2 \times 10^{-6} \mathrm{M} \mathrm{HCl}\):\( \mathrm{pH} = -\log(2 \times 10^{-6}) \approx 5.7 \)Since this value is less than 6, it is excluded.2. For \(10^{-8} \mathrm{M} \mathrm{HCl}\):\( \mathrm{pH} = -\log(10^{-8} + 10^{-7}) \approx 7 - 8\), approximate based on pure water and proton concentrations makes it approach neutral pH.Only the second acid sample fits in the 6-7 pH range.

Step 4: Result Verification

From calculations:- Solution 1 (\(2 \times 10^{-6} \mathrm{M} \mathrm{NaOH}\)) gives pH of \( \approx 8.3 \) and is not between 6 and 7.- Solution 2 (\(2 \times 10^{-6} \mathrm{M} \mathrm{HCl}\)) gives pH of \( \approx 5.7 \) and is not between 6 and 7.- Solution 3 (\(10^{-8} \mathrm{M} \mathrm{HCl}\)) gives pH closer to pure water's due to dilution, near 7.- Solution 4 (\(10^{-13} \mathrm{M} \mathrm{NaOH}\)) primarily reflecting neutral pH due to water equilibrium, remains near 7.Thus, solutions 3 and 4 fit the criteria.

Key Concepts

Strong Acids and Bases

Strong acids and bases are highly effective at changing the pH of a solution. They completely dissociate in water, meaning they split into their constituent ions. For instance, a strong acid like hydrochloric acid (HCl) will dissociate into hydrogen ions (H\(^+\)) and chloride ions (Cl\(^-\)). On the other hand, a strong base such as sodium hydroxide (NaOH) dissociates into sodium ions (Na\(^+\)) and hydroxide ions (OH\(^-\)).

When you are dealing with strong acids and bases, it's important to realize that their dissociation is so complete that the concentration of the acid or base is equal to the concentration of the resulting ions.

• Strong acids have a pH less than 7 because they increase the concentration of hydrogen ions in solution.
• Strong bases have a pH greater than 7 as they increase the concentration of hydroxide ions.

This complete dissociation makes calculations straightforward because there's no need to consider any undissociated molecules.

Accurate pH calculation is crucial to understanding how acidic or basic a solution is, and it's a key element in various scientific and industrial processes.

Kw - Ionic Product of Water

The ionic product of water, denoted as K\(_w\), is a fundamental concept in understanding the chemistry of aqueous solutions. It is the product of the concentrations of hydrogen ions and hydroxide ions in water at a specific temperature, usually 25°C, and its value is 10\(^{-14}\) mol\(^2\)/L\(^2\).

This equilibrium constant is expressed as:
\[ K_w = [H^+][OH^-] = 10^{-14} \]
Knowing the value of K\(_w\) allows us to calculate the concentration of hydrogen ions if we know the concentration of hydroxide ions, and vice versa.

• In pure water, the concentrations of hydrogen ions and hydroxide ions are equal, thus each is 10\(^{-7}\) M, resulting in a neutral pH of 7.
• An increase in hydrogen ions leads to a decrease in hydroxide ions and a lower pH, indicating acidity, whereas an increase in hydroxide ions does the opposite.

Understanding K\(_w\) helps us reason out the acid-base properties in solutions, particularly in calculations involving low concentrations, where deviations from expected pH values might occur due to ion interactions with water.

Dissociation of NaOH and HCl

Dissociation refers to the process by which compounds split into their constituent ions when dissolved in water. For substances like NaOH and HCl, this occurs completely due to their strong acidic and basic nature.

When NaOH is dissolved in water, it dissociates as follows:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
The production of OH\(^-\) ions makes the solution basic, typically resulting in a pH greater than 7.

Similarly, when HCl is dissolved, it dissociates as follows:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
This increases the concentration of H\(^+\) ions, resulting in an acidic solution with a pH less than 7.

• For strong acids and bases, the concentration of the initial compound directly corresponds to the concentration of the ions.
• Complete dissociation simplifies calculations as it assumes that the initial concentration translates entirely into ionic concentrations.

Understanding the dissociation helps in predicting the behavior of solutions, their pH, and adjusting conditions in processes like titration or neutralization.