Q.38 The concentration of OH– in a solution with H+
concentration of 1.3 × 10-4 M is:
- 7.7 × 10-22 M
- 7.0 × 10-12 M
- 7.7 × 10-11 M
- 1 × 10-14 M
Concentration of OH– in Solution with H+ 1.3 × 10-4 M: Detailed Calculation Guide
The correct answer is 7.7 × 10-11 M, calculated using the ionic product of water, Kw = 1.0 × 10-14 at 25°C, where [OH–] = Kw/[H+].
Step-by-Step Solution
In aqueous solutions, water dissociates as H2O ⇌ H+ + OH–, with the ion product Kw = [H+][OH–] = 1.0 × 10-14 (at 25°C).
Given [H+] = 1.3 × 10-4 M, solve for [OH–]:
[OH–] = 1.0 × 10-14
1.3 × 10-4
[OH–] = (1.0 × 10-14) / (1.3 × 10-4) = (1/1.3) × 10-10 ≈ 0.77 × 10-10 = 7.7 × 10-11 M
This confirms the solution is acidic since [H+] > [OH–].
Option Analysis
- 7.7 × 10-22 M: Far too low; this might result from incorrect exponent handling, like 10-14 × 10-4 × 1.3, which is erroneous.
- 7.0 × 10-12 M: Close but inaccurate; approximates 10-14/10-3 = 10-11, ignoring the 1.3 factor precisely.
- 7.7 × 10-11 M: Correct, as derived above using exact division.
- 1 × 10-14 M: Equals Kw itself, a common mistake confusing the product with [OH–] in neutral water.
The concentration of OH- in a solution with H+ concentration of 1.3 × 10-4 M is a fundamental chemistry problem testing the ionic product of water, Kw. This guide explains the exact calculation, why 7.7 × 10-11 M is correct, and analyzes all options for exam prep like JEE or NEET.
Core Concept: Ionic Product of Water
Water’s self-ionization gives Kw = [H+][OH–] = 1.0 × 10-14 at 25°C, constant across solutions.
In acidic conditions like this (pH = -log(1.3 × 10-4) ≈ 3.89), [H+] exceeds [OH–], but their product stays fixed.
Precise Calculation
[OH–] = 1.0 × 10-14
1.3 × 10-4
Use scientific notation division: 10-14 / 10-4 = 10-10, then divide by 1.3 (≈0.769) to get 7.7 × 10-11.
Why Other Options Fail
| Option | Value | Reason for Incorrectness |
|---|---|---|
| A | 7.7 × 10-22 M | Wrong operation (multiplication instead of division). |
| C | 7.7 × 10-11 M | Correct per Kw formula. |
| B | 7.0 × 10-12 M | Rough approx for [H+]=10-3, not 1.3×10-4. |
| D | 1 × 10-14 M | Confuses Kw with [OH–] in neutral solution. |
Exam Tips
Practice with logs for pOH: pH ≈3.89, pOH=14-3.89=10.11, [OH–]=10-10.11≈7.7×10-11. Always verify units (M) and temperature for Kw.
