Q.46 The [H⁺] of 0.1 N acetic acid solution is 1.33 × 10⁻³. The pH of the solution (correct to two decimal places) is ________.

The pH of the 0.1 N acetic acid solution is calculated directly from the given [H⁺] concentration using the standard pH formula. This yields a value of 2.88 when rounded to two decimal places.​

pH Calculation

The pH is defined as the negative base-10 logarithm of the hydrogen ion concentration: pH=−log⁡10[H+]pH=−log10[H+]. Substituting the given [H⁺] = 1.33×10−31.33×10−3 M gives pH=−log⁡10(1.33×10−3)pH=−log10(1.33×10−3). This computes to approximately 2.876, which rounds to 2.88 correct to two decimal places.​

Context for Acetic Acid

Acetic acid (CH₃COOH) is a weak acid with typical Ka≈1.8×10−5Ka≈1.8×10−5, leading to partial dissociation in 0.1 N (equivalent to 0.1 M) solution. The provided [H⁺] = 1.33×10−31.33×10−3 M matches expected equilibrium values from the dissociation CH3COOH⇌H++CH3COO−CH3COOH⇌H++CH3COO−, where [H+]≈Ka⋅C≈1.34×10−3[H+]≈Ka⋅C≈1.34×10−3 M. No options are provided in the query, so the direct numerical answer applies without elimination.​

Introduction to pH of 0.1 N Acetic Acid [H⁺] 1.33 × 10⁻³ Calculation

Determining the pH of 0.1 N acetic acid solution is a core concept in acid-base chemistry, especially for competitive exams like CSIR NET Life Sciences. When given [H⁺] = 1.33 × 10⁻³ M, the calculation simplifies to a direct application of the pH formula, yielding pH = 2.88 (correct to two decimal places). This pH of 0.1 N acetic acid [H⁺] 1.33 × 10⁻³ problem tests understanding of weak acid dissociation and logarithmic math.​

Step-by-Step pH Calculation Method

• Formula: pH=−log⁡10[H+]pH=−log10[H+].​

• Plug in values: pH=−log⁡10(1.33×10−3)pH=−log10(1.33×10−3).​

  • log⁡10(1.33×10−3)=log⁡101.33+log⁡1010−3=0.1239−3=−2.8761log10(1.33×10−3)=log101.33+log1010−3=0.1239−3=−2.8761.

  • Thus, pH = 2.876 ≈ 2.88.​

• Verification: For 0.1 N CH₃COOH (Ka≈1.8×10−5Ka≈1.8×10−5), [H⁺] ≈ 1.8×10−5×0.1=1.34×10−31.8×10−5×0.1=1.34×10−3 M, confirming the given value.​

Why No Options? Direct Numerical Solution

This numerical response-type question (common in CSIR NET) has no multiple-choice options to evaluate. Common distractors in similar problems include:

• Misrounding to 2.87 or 2.9 (ignores two-decimal precision).​

• Assuming strong acid full dissociation (pH = 1, incorrect for weak acid).​

• Forgetting negative log (yielding 2.88 instead of -2.88).​
  The correct approach always uses the given [H⁺] directly.​

Applications in Biochemistry and Exam Prep

In biochemistry, acetic acid pH (around 2.88) affects enzyme activity and buffers like acetate. For CSIR NET, master ICE tables for weak acids: Ka=x2/C−x≈x2/C. Practice similar problems to ace quantitative sections.​