Quick Summary
Concept Tested: pH of Dilute Strong Acids & Water Autoionization
Chapter: Chemical Equilibrium
Difficulty: ★★★★ (Medium)
Time: 1-2 mins
Key Formula: $$K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14}$$
Answer: B
Reasoning: The statement in Option (B) is false because the pH of $1 \times 10^{-8} \text{ M HCl}$ is approximately 6.96, not 8. The contribution of water’s autoionization ($10^{-7} \text{ M}$) cannot be neglected in such a dilute solution.
The Question
Which one of the following statements is not true? [2003]
(A) pH + pOH = 14 for all aqueous solutions
(B) The pH of $1 \times 10^{-8} \text{ M HCl}$ is 8
(C) 96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of Cu
(D) The conjugate base of H3PO4 is HPO42-
Quick Answer
The correct answer is Option (B).
The statement “The pH of $1 \times 10^{-8} \text{ M HCl}$ is 8” is not true. Due to the autoionization of water, the actual hydrogen ion concentration is higher ($1.1 \times 10^{-7} \text{ M}$), resulting in a pH of approximately 6.96, which is slightly acidic.
Why Other Options Are Incorrect
Option (A): pH + pOH = 14 for all aqueous solutions
This statement is true at 25°C. It is derived from the definition of $K_w$ (the ion product of water). Since $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$, taking the negative logarithm of both sides gives $\text{pH} + \text{pOH} = 14$. This relationship holds for any aqueous solution at this temperature.
Option (C): 96,500 coulombs of electricity when passed through a CuSO4 solution deposits 1 gram equivalent of Cu
This statement is true based on Faraday’s laws of electrolysis. 96,500 coulombs (1 Faraday) deposits 1 gram-equivalent of any substance. For copper, the reduction half-reaction is $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$. Since 1 mole of electrons (1F) deposits 1 mole of charge, and 2 moles of electrons (2F) deposit 1 mole of Cu, 1F deposits 1 gram-equivalent (31.75 g) of Cu.
Option (D): The conjugate base of H3PO4 is HPO42-
This statement is true. Phosphoric acid ($\text{H}_3\text{PO}_4$) is a triprotic acid. Removing one proton ($\text{H}^+$) gives $\text{H}_2\text{PO}_4^-$, removing a second gives $\text{HPO}_4^{2-}$ (the conjugate base of the second dissociation), and removing a third gives $\text{PO}_4^{3-}$.
Video Solution
Video Solution Coming Soon
Understanding the Concept
The core concept here is the autoionization of water and the definition of pH. In pure water at 25°C, the concentration of $\text{H}^+$ ions is $10^{-7} \text{ M}$, giving a pH of 7. When a strong acid like HCl is added, it increases the $\text{H}^+$ concentration. However, in extremely dilute solutions (like $10^{-8} \text{ M}$), the contribution of $\text{H}^+$ from the autoionization of water ($10^{-7} \text{ M}$) becomes significant. You cannot simply ignore the water’s ions; you must add them to the acid’s contribution to find the total concentration.
Detailed Step-by-Step Solution
Step 1: Analyze the Autoionization of Water
In any aqueous solution, water molecules undergo autoionization to produce $\text{H}^+$ and $\text{OH}^-$ ions. The equilibrium constant for this process is the ion product of water ($K_w$).
$$\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$$
At 25°C, $K_w = 1 \times 10^{-14}$. Since the dissociation is equal, the concentration of $\text{H}^+$ from water is always $10^{-7} \text{ M}$ in pure water, and it remains a baseline value that cannot be ignored in very dilute solutions.
Step 2: Calculate the Total [H+]
HCl is a strong acid, meaning it dissociates completely. The problem gives the concentration of HCl as $1 \times 10^{-8} \text{ M}$. We must sum the $\text{H}^+$ contributed by the HCl with the $\text{H}^+$ contributed by the water.
$$[\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{from HCl}} + [\text{H}^+]_{\text{from water}}$$
$$[\text{H}^+]_{\text{total}} = 1 \times 10^{-8} \text{ M} + 1 \times 10^{-7} \text{ M}$$
$$[\text{H}^+]_{\text{total}} = 1 \times 10^{-8} + 10 \times 10^{-8} = 11 \times 10^{-8} \text{ M}$$
$$[\text{H}^+]_{\text{total}} = 1.1 \times 10^{-7} \text{ M}$$
Step 3: Calculate the pH
The pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration.
$$\text{pH} = -\log [\text{H}^+]$$
Substitute the total concentration found in Step 2:
$$\text{pH} = -\log (1.1 \times 10^{-7})$$
Using the logarithm property $\log(ab) = \log a + \log b$:
$$\text{pH} = -(\log 1.1 + \log 10^{-7})$$
$$\text{pH} = -(0.0414 – 7)$$
$$\text{pH} = 6.9586$$
Rounding to two decimal places, the pH is 6.96.
Final Answer
✔ Option (B) is the correct choice because the statement “The pH of $1 \times 10^{-8} \text{ M HCl}$ is 8” is false. The correct pH is approximately 6.96.
Essential Formulas for This Topic
- Ion Product of Water: $$K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \text{ at } 25^\circ\text{C}$$
- Definition of pH: $$\text{pH} = -\log [\text{H}^+]$$
- Definition of pOH: $$\text{pOH} = -\log [\text{OH}^-]$$
- Relationship: $$\text{pH} + \text{pOH} = 14$$
- Hydroxide Concentration from pH: $$[\text{OH}^-] = \frac{1 \times 10^{-14}}{[\text{H}^+]}$$
Common Mistakes to Avoid
Mistake 1: Ignoring Water’s Contribution
Wrong Thinking: Calculating pH directly as $-\log(10^{-8}) = 8$.
Correct Approach: Always check if the acid/base concentration is less than $10^{-6} \text{ M}$. If it is, add $10^{-7} \text{ M}$ (from water) to the acid concentration before calculating pH.
Mistake 2: Assuming pH > 7 means Basic
Wrong Thinking: Thinking a pH of 6.96 is basic.
Correct Approach: pH < 7 is always acidic, regardless of how close it is to 7. The pH scale is continuous, not discrete.
Mistake 3: Forgetting the Temperature Factor
Wrong Thinking: Assuming $K_w$ is always $1 \times 10^{-14}$.
Correct Approach: $K_w$ increases with temperature. At higher temperatures, the pH of pure water is less than 7. However, for standard JEE questions, assume 25°C unless stated otherwise.
Key Concept Summary
- Water autoionization provides a baseline of $10^{-7} \text{ M} \text{ H}^+$.
- In very dilute solutions ($< 10^{-6} \text{ M}$), water's contribution is significant.
- The sum of acid/base concentration and water’s contribution equals total $[\text{H}^+]$.
- pH < 7 indicates an acidic solution.
Golden Rule: Never calculate the pH of a strong acid or base without checking if the concentration is lower than $10^{-6} \text{ M}$. If it is, you must add $10^{-7} \text{ M}$ to the concentration.
Frequently Asked Questions
Q: What is the pH of $1 \times 10^{-7} \text{ M HCl}$?
A: The pH would be 7. This is because $[\text{H}^+]_{\text{total}} = 10^{-7} + 10^{-7} = 2 \times 10^{-7} \text{ M}$, so $\text{pH} = -\log(2 \times 10^{-7}) \approx 6.7$.
Q: Can the pH of a solution be negative?
A: Yes. For very concentrated strong acids (e.g., $1 \text{ M HCl}$), $[\text{H}^+] = 1$, so $\text{pH} = -\log(1) = 0$. For $10 \text{ M HCl}$, $\text{pH} = -1$.
Q: Why does water autoionize?
A: Water molecules are polar. Occasionally, a water molecule can transfer a proton to another water molecule, resulting in a hydronium ion ($\text{H}_3\text{O}^+$) and a hydroxide ion ($\text{OH}^-$).
Q: Does the pH scale end at 14?
A: No. The pH scale is theoretically unbounded. It can go below 0 for very acidic solutions and above 14 for very basic solutions.
Prerequisites to Solve This Question
- Understanding the definition of pH and pOH.
- Knowledge of the strong acid dissociation (HCl $\rightarrow \text{H}^+ + \text{Cl}^-$).
- Understanding the ion product of water ($K_w$).
- Familiarity with logarithmic calculations.
After Solving This, You Can:
- ✔ Calculate the pH of extremely dilute strong acids.
- ✔ Understand the role of water’s autoionization in equilibrium calculations.
- ✔ Differentiate between “acidic” and “neutral” based on precise values.
Study Tips for This Topic
Focus on memorizing the value of $K_w$ at 25°C ($10^{-14}$). Practice solving problems involving $10^{-8} \text{ M}$ HCl and $10^{-8} \text{ M}$ NaOH to solidify the concept that $10^{-8} \text{ M}$ NaOH is actually acidic (pH ~6.96) and $10^{-8} \text{ M}$ HCl is slightly acidic (pH ~6.96), while $10^{-7} \text{ M}$ HCl is neutral (pH ~7).
Difficulty Rating & Exam Frequency
Difficulty: ★★★★ (Medium-Hard)
JEE Main Frequency: Moderate. Often appears in the numerical type or conceptual type questions in the Chemistry section.
Importance: High. This concept is foundational for understanding acid-base buffers and solubility equilibria.
Related Questions from Chemical Equilibrium
Written by Nishant Kumar Gupta
Founder of Padho Likho JEE & Senior Chemistry Educator — 12+ Years Experience, Ex-Faculty Allen/Aakash/Narayana.
Last Updated: July 2026
Question Source: JEE Main 2003 PYQ
Topic: Chemical Equilibrium