Question

What is the pH of a solution prepared by adding 1.46g of NaNO₂ to 143.00 mL of water? (Base Ionization Constant for NaNO₂ = 2.22E-11)

Answer 1

Molar mass of NaNO2,
MM = 1*MM(Na) + 1*MM(N) + 2*MM(O)
= 1*22.99 + 1*14.01 + 2*16.0
= 69 g/mol


mass(NaNO2)= 1.46 g

use:
number of mol of NaNO2,
n = mass of NaNO2/molar mass of NaNO2
=(1.46 g)/(69 g/mol)
= 2.116*10^-2 mol
volume , V = 1.43*10^2 mL
= 0.143 L


use:
Molarity,
M = number of mol / volume in L
= 2.116*10^-2/0.143
= 0.148 M

NO2- dissociates as:

NO2- +H2O     ----->     HNO2   +   OH-
0.148                   0         0
0.148-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 ((2.22*10^-11)*0.148) = 1.813*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.813*10^-6 M



So, [OH-] = x = 1.813*10^-6 M


use:
pOH = -log [OH-]
= -log (1.813*10^-6)
= 5.7417


use:
PH = 14 - pOH
= 14 - 5.7417
= 8.2583
Answer: 8.26