40 mL sample of 0.1 M solution of nitric acid is added to 20 mL of 0.3 M aqueous ammonia. What is the pH of the resulting solution? ( p K b = 4.7447 )

Question

40 mL sample of 0.1 M solution of nitric acid is added to 20 mL of 0.3 M aqueous ammonia. What is the pH of the resulting solution? ( p K b = 4.7447 ), 8.95, 89.5, 0.895, 4.48

MARKS App · Accepted Answer

H N O 3 4 m m o l + N H 3 6 m m o l → N H 4 N O 3 4 m m o l [Since W B ( N H 3 ) is left and salt ( N H 4 N O 3 ) is formed so it is a basic buffer] Base left = 6 – 4 = 2 mmol [Acid] = 40 × 0.1 = 4 mmol [Base] = 20 × 0.3 = 6 pOH = p K b + l o g 10 [ S a l t ] [ B a s e ] So, pOH = 4.7447 + l o g 4 / 2 = 5.0457 and pH = 14 – 5.0457 = 8.95.

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