Question

1. The pH of an aqueous solution of 0.257 M potassium nitrite, KNO₂ (aq), is _____.

This solution is (acidic, basic, neutral)

2. The pH of an aqueous solution of 0.285 M ammonium iodide, NH₄I (aq), is _____.

This solution is (acidic, basic, neutral)

Answer 1

1)

Ka of HNO2 = 4.5*10^-4

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/4.5*10^-4

Kb = 2.222*10^-11

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.257 0 0

0.257-x x x

Kb = [HNO2][OH-]/[NO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.222*10^-11)*0.257) = 2.39*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.39*10^-6 M

use:

pOH = -log [OH-]

= -log (2.39*10^-6)

= 5.6216

use:

PH = 14 - pOH

= 14 - 5.6216

= 8.38

since pH is more than 7, it is basic

Answer: 8.38, basic

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