Question

15. What is the pH of solution made by adding 3.9 g NaNH, into water to make a 500 ml
solution K (NH,) = 2 x 10^ [Na = 23, N = 14, H = 1]
(A) 13.3
(B) 0.7 (C) 5.3 (D) 13.7​

Answer 1

Answer:

a)13.3

Explanation:

According to given data, number of moles of NaNH2

[NaNH2]= (3.9/39) × (500/1000) = 0.2M

So pH of a solution of strong base-weak acid salt

pH = 7 + (1/2) × pKa + (1/2) × log C

By putting given values,

pH = 7 + (1/2) × (14 - (5−log 2) ) + (1/2)×log 0.2= 13.3