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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
- The wrong statements about glycerol is
- Standard reduction potential for Al³⁺/Al , Fe²/Fe, Br/Br are ₋1.66V, ₋0.45V, 1.09V
- In the extraction of aluminium from bauxite, cryolite is used to
- The rate of reactions exhibiting negative activation energy
- In Kjeldahl’s method,ammonia from 5g of food neutralizes 30 cm3 of 0.1 N acid.
Topics: Equilibrium (104)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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