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Q.

pKaCH3COOH is 4.74.x moles of lead acetate and 0.1 mole of acetic acid in 1 L solution make a solution of pH = 5.04. Hence, x is

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a

0.2

b

0.05

c

0.1

d

0.02

answer is C.

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Detailed Solution

xmolCH3COO2Pb=2xmolCH3COO ∴ CH3COOH=0.1M CH3COO−=2×M ∴ pH=pKa+log⁡CH3COO−CH3COOH 5.04=4.74+log⁡20x⇒20x=2 x=0.1mol
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