A 0.1 M solution of a weak monobasic acid (HA) is titrated with 0.1 M NaOH. Given that the $K_a$ of HA is $1.8 \times 10^{-5}$, what is the approximate pH at the half-equivalence point?

## CORE INFORMATION\n✓ **Answer**: B\n## CONCEPT FRAMEWORK\n **Primary Concept**: Acid-base titrations and the Henderson-Hasselbalch equation.\n **Related Topics**: Weak acids, pH calculations, equivalence point.\n **Prerequisites**: Understanding of $K_a$, $pK_a$, and the Henderson-Hasselbalch equation.\n## SOLUTION PATHWAY\n**Given Information**:\n $[HA] = 0.1 M$\n $[NaOH] = 0.1 M$\n $K_a = 1.8 \times 10^{-5}$\n At half-equivalence point, $[HA] = [A^-]$ (conjugate base)\n**Method & Steps**:\n 1. Calculate $pK_a$:\n Reasoning: The $pK_a$ is the negative logarithm of the acid dissociation constant ($K_a$).\n Formula/Rule: $pK_a = -log_{10}(K_a)$\n 2. Apply the Henderson-Hasselbalch equation:\n Reasoning: At the half-equivalence point, the concentrations of the weak acid (HA) and its conjugate base ($A^-$) are equal. The Henderson-Hasselbalch equation simplifies significantly in this scenario.\n Formula/Rule: $pH = pK_a + log_{10} \frac{[A^-]}{[HA]}$\n Working: Since $[A^-] = [HA]$ at the half-equivalence point, the $log_{10} \frac{[A^-]}{[HA]}$ term becomes $log_{10}(1) = 0$. Therefore, $pH = pK_a$.\n 3. Calculate the pH:\n Working: $pK_a = -log_{10}(1.8 \times 10^{-5}) \approx 4.74$\n Result: $pH \approx 4.74$\n Verification: The pH at the half-equivalence point should be equal to the $pK_a$ for a weak acid titration. Our result confirms this.\n ## OPTION ANALYSIS\nA :x: Incorrect. This value is significantly lower than the expected $pK_a$.\nB ✓: Correct. This value matches the calculated $pK_a$, which is the pH at the half-equivalence point.\nC :x: Incorrect. This is the neutral pH, not expected for a weak acid titration.\nD :x: Incorrect. This value is far too high, indicating a basic solution, which is not the case at the half-equivalence point.\n ## LEARNING AIDS\n **Common Mistakes**:\n 1.Forgetting to use the negative logarithm when calculating $pK_a$ from $K_a$: Always remember that $pK_a = -log_{10}(K_a)$.\n 2.Misapplying the Henderson-Hasselbalch equation: Ensure you understand the conditions under which the equation simplifies (e.g., at the half-equivalence point).\n **Quick Tips**:\n 1.At the half-equivalence point of a weak acid titration, the pH is directly equal to the $pK_a$ of the acid. This is a useful shortcut for quick estimations.\n 2.Remember that $pK_a$ represents the acidity of the weak acid; a lower $pK_a$ implies a stronger acid.

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A 0.1 M solution of a weak monobasic acid (HA) is titrated with 0.1 M NaOH. Given that the $K_a$ of HA is $1.8 imes 10^{-5}$ , what is the approximate pH at the half-equivalence point?

2.37

4.74

7.00

9.26

Question Tags

📚 Subject:Chemistry

📑 Chapter:Organic Chemistry: Some basic principles and techniques

🎯Topic:Qualitative & Quantitative Analysis

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :2 mins

🔄Ncert Concepts :Titration, Weak Acid, pH, pKa, Half-Equivalence Point

A mixture of two volatile liquids A and B is distilled. The boiling point of A is 70°C and that of B is 90°C. Assuming ideal behavior, which of the following statements is true regarding the initial distillate?

The distillate will be richer in A than the original mixture.

The distillate will be richer in B than the original mixture.

The distillate will have the same composition as the original mixture.

The distillate will contain only A.

Question Tags

📚 Subject:Chemistry

📑 Chapter:Organic Chemistry: Some basic principles and techniques

🎯Topic:Qualitative & Quantitative Analysis

🎓 Ncert Level:Understanding

🎯Difficulty :Hard

⚡ Time Required :1.5 mins

🔄Ncert Concepts :Distillation, Boiling Point, Raoult's Law

A sample is analyzed using paper chromatography. Component X has an Rf value of 0.6, while component Y has an Rf value of 0.3. Which component is more strongly adsorbed onto the stationary phase?

Component X

Component Y

Both components are equally adsorbed.

Cannot be determined from Rf values.

Question Tags

📚 Subject:Chemistry

📑 Chapter:Organic Chemistry: Some basic principles and techniques

🎯Topic:Qualitative & Quantitative Analysis

🎓 Ncert Level:Application

🎯Difficulty :Hard

⚡ Time Required :1 mins

🔄Ncert Concepts :Chromatography, Rf Value, Adsorption

In the Kjeldahl method for nitrogen estimation, a sample containing 0.42 g of an organic compound yielded ammonia which was neutralized by 20 mL of 1 M $H_2SO_4$ . The percentage of nitrogen in the compound is: