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40 mL of 0.1 M ammonia solution is mixed with 20 mL of 0.1 M HCl. What is the pH of the mixture? ( pKb of ammonia solution is 4.74)
Options
(a) 4.74
(b) 2.26
(c) 9.26
(d) 5
Correct Answer:
9.26
Explanation:
40mL of 0.1 M ammonia solution = 40×0.1
= 4 milli equivalent ammonia solution,
20 mL of 0.1 M HCl = 20×0.1,
= 2 milliequivalent of HCl,
NH₄OH + HCl → NH₄Cl + H₂O. Initial milli.eqv. 4 2 0. Milli-eqv.after 4-2 0 2 reaction = 2. Therefore pOH = pK(b) + log[NH₄Cl]/NH₄OH] = 4.74 + log2/2
= 4.74 + log 1 = 4.74.
Therefore pH = 14 – 4.74 = 9.26.
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Topics: Equilibrium
(104)
Subject: Chemistry
(2512)
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