Problem 10 - A Complete Resource Book in Chemistry for JEE Main

Question

The \(\mathrm{pH}\) of \(10^{-10} \mathrm{M} \mathrm{NaOH}\) solution is (a) 10 (b) \(7.01\) (c) \(6.99\) (d) 4

Step-by-Step Solution

Answer

The correct answer is (c) 6.99.

Step 1: Understanding the Solution

Realize that a sodium hydroxide (NaOH) solution is a basic solution, and it contributes ext{OH}^- ions in the solution.

Step 2: Calculate the Contribution of NaOH

Given the concentration of ext{NaOH} is \(10^{-10} \, \text{M}\), the concentration of ext{OH}^- ions contributed by NaOH is \(10^{-10} \, \text{M}\).

Step 3: Consider the Contribution of Water

Water also dissociates into ext{H}^+ and ext{OH}^- ions. In pure water at 25°C, these concentrations are both \(10^{-7} \, \text{M}\). The total ext{OH}^- concentration in the NaOH solution will be the sum of the contribution from NaOH and water.

Step 4: Calculate Total OH^- Concentration

The total ext{OH}^- concentration is the sum of \(10^{-10} \, \text{M}\) from NaOH and \(10^{-7} \, \text{M}\) from water, which gives \(10^{-7} \, \text{M}\) because it's the larger contribution.

Step 5: Calculate pOH

The \(\text{pOH}\) is calculated as \(-\log{[\text{OH}^-]} = -\log{10^{-7}} = 7\).

Step 6: Calculate pH from pOH

Since \(\text{pH} + \text{pOH} = 14\), then \(\text{pH} = 14 - 7 = 7\).

Step 7: Conclusion

Given the calculations, the \(\text{pH}\) of the solution is just slightly higher than neutral (7), therefore it is approximately \(6.99\).

Key Concepts

Concentration of Ions

In a solution, the concentration of ions refers to the amount of dissolved ions present in a given volume of liquid. For computing the concentration of ions, you usually consider eitherthe amount of a particular ion given directly by a solute, like sodium hydroxide (NaOH) in an aqueous solution, or from water itself through its natural process of dissociation.
NaOH is a source of hydroxide ions (OH⁻) because it dissociates completely in water. In the given problem, the concentration of NaOH is very low, at only \(10^{-10} \text{ M}\), resulting in a corresponding OH⁻ ion concentration of \(10^{-10} \text{ M}\). This is crucial for understanding how much this specific ion contributes to the overall chemistry of the solution.
Remember that each molecule of NaOH gives one OH⁻ ion upon dissolution. Thus, the concentration of NaOH directly equals the concentration of OH⁻ ions from it.

Water Dissociation

Water dissociation is a natural process where water (H₂O) splits into hydrogen ions (H⁺) and hydroxide ions (OH⁻). Even without any added solute, pure water undergoes this ionization, albeit at very low concentrations.
At 25°C, these concentrations are both \(10^{-7} \text{ M}\). Therefore, in any aqueous solution, even without additional substances, there will always be OH⁻ and H⁺ ions present due to this dissociation.
In the context of the provided NaOH solution, the water itself provides its own OH⁻ ions, adding to the concentration coming from the sodium hydroxide. Since \(10^{-7} \text{ M}\) is a larger concentration than the \(10^{-10} \text{ M}\) from NaOH, water's contribution becomes the dominant factor in the total OH⁻ concentration.

pOH Calculation

The \(\text{pOH}\) is a measure of the hydroxide ion concentration in a solution, calculated as the negative logarithm of the OH⁻ concentration. To find the pOH in our NaOH solution, it is crucial to first have the total OH⁻ concentration in mind.
Once the primary OH⁻ source concentrations—including those from water and NaOH—are summed up, you proceed by calculating the pOH. For our example, the \(\text{pOH}\) is given by \(-\log{[\text{OH}^-]}\).
Because water supplies more significant OH⁻ with \(10^{-7} \text{ M}\) compared to NaOH, the calculation simplifies to \(-\log{10^{-7}} = 7\). The pOH provides insight into the basicity of a solution, being inversely related to the hydroxide ion concentration.

Sodium Hydroxide Solution

Sodium hydroxide (\(\text{NaOH}\)) is known as a strong base because it dissociates completely in water, meaning each NaOH unit releases a hydroxide ion (OH⁻) into the solution. Surpassed only by very concentrated acids, NaOH solutions can be hazardous and must be handled with care but are widely used in chemistry labs to demonstrate basic solutions.
Even when at mid to low molarities, NaOH solutions affect the pH of the water they are dissolved in, elevating it above neutral. Despite its complete dissociation tendencies, in our given problem, the particularly low concentration (\(10^{-10} \text{ M}\)) results majorly in contribution to the basic environment coming naturally from water (
not from the NaOH itself). It is a fascinating showcase of how even a strong base can sometimes not significantly affect pH due to concentration conditions.