Q.26 The ionic product of water at 40 °C is 2.92 ×10-14 M². The pH of water at 40 °C is ___________ (rounded off to 2 decimal places).
The pH of Pure Water at 40°C
The pH of pure water at 40°C, where the ionic product Kw = 2.92 × 10-14 M², is 6.77. This value is calculated from the autoionization equilibrium where [H⁺] = [OH⁻] = √Kw, and differs from 7.00 at 25°C because Kw increases with temperature.
Calculation Steps
Pure water autoionizes as H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻]. In neutral water, [H⁺] = [OH⁻], so:
[H⁺] = √(2.92 × 10-14) = 1.71 × 10-7 M
Then, pH = -log₁₀([H⁺]) = -log₁₀(1.71 × 10-7) = 6.77 (rounded to two decimal places). Water remains neutral at this pH since [H⁺] = [OH⁻], despite the value below 7.
Temperature Effect on Kw
At 25°C, Kw = 1.00 × 10-14, yielding pH 7.00. Higher temperatures increase Kw due to the endothermic autoionization reaction, shifting equilibrium rightward per Le Châtelier’s principle. Thus, neutral pH decreases above 25°C.
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Introduction to Ionic Product of Water at 40°C
The ionic product of water at 40°C, denoted Kw = 2.92 × 10-14 M², governs autoionization and pH in aqueous solutions. For CSIR NET Life Sciences aspirants, understanding this temperature-dependent value is crucial for acid-base equilibrium questions. Unlike 25°C where neutral pH is 7.00, elevated temperatures alter neutrality, making precise calculations essential.
Detailed pH Calculation Method
Start with Kw = [H⁺][OH⁻] = 2.92 × 10-14. In pure water, [H⁺] = [OH⁻] = √Kw.
- Compute square root:
√(2.92 × 10-14) = 1.71 × 10-7 M - Apply pH formula:
pH = -log(1.71 × 10-7) = 6.77
This matches exam expectations, rounded to two decimal places.
Why pH Changes with Temperature
Autoionization 2H₂O ⇌ H₃O⁺ + OH⁻ is endothermic, so Kw rises from 1 × 10-14 at 25°C to 2.92 × 10-14 at 40°C. Neutral point remains where [H⁺] = [OH⁻], but pH drops below 7—key for interpreting “neutrality” in exams.
CSIR NET Exam Relevance
Competitive exams test this via direct calculations or Kw trends. Common pitfalls include:
- Assuming pH = 7 regardless of temperature
- Misapplying log rules
- Confusing ionic product with ion product
Practice Question: At 50°C, Kw ≈ 5.5 × 10-14, pH ≈ 6.78. Master for scoring in physical chemistry sections.
Key Takeaway: pH of pure water at 40°C = 6.77
