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When 10 mL of 0.1 M acetic acid (pKa = 5.0 ) is titrated against 10mL of 0.1 M ammonia solution (pKb = 5.0),the equivalence point occurs at pH
Options
(a) 5
(b) 6
(c) 7
(d) 9
Correct Answer:
7
Explanation:
pKₐ = -logKₐ : pK(b) = -logK(b),
pH = -1/2[logKₐ + log K(w) – logK(b),
-1/2[-5 + log(1*10⁻¹⁴)-(-5)],
-1/2[-5-14+5]=-1/2(-14)=7.
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Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In the brown ring test for the nitrate ion, the brown colour is due to
- What is the pH of 0.01 M glycine solution? For glycine, Ka₁ = 4.5 ˣ 10⁻³
- A mixture of two salts is not soluble in water but dissolves completely in dilu
- The element that does not form a monoxide is
- The cylindrical shape of an alkyne is due to
Topics: Equilibrium (104)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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