Question

# Calculate the pH of 300 mL of a 0.30 M aqueous solution of sodium nitrite (NaNO2)...

Calculate the pH of 300 mL of a 0.30 M aqueous solution of sodium nitrite (NaNO2) at 25 °C given that the Ka of nitrous acid (HNO2) is 4.5×10-4.

A) 5.59

C) 4.46

D) 7.00

E) 12.87

a)

pH of a M = 0.30 of a SALT formed from WEAK acid; Strong base

in solution, this will form

Na+ and NO2- in solution

NO2- is the conjugate base of a weak acid, HNO2

it will form an equilibrium

NO2- + H2O <-> HNO2+ OH-

there are OH- in solution, so this will be BASIC

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

Ka for HCN = 4.5*10^-4

we need Kb, so:

Kw = KaKb

Kw = 10^-14 always for water at 25°C

then

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

Kb = (10^-14)/(4.5*10^-4) = 2.22222*10^-11

now

Assume, in equilibrium that

[OH-] = [HNO2] = x

and

[NO2-] = 0.3 - x

then, substitute all in Kb expression

Kb = [HA][OH-]/[A-]

(2.22222*10^-11) = (x*x)/(0.3 - x)

solve for x using quadratic equation

x = [OH-] = 2.58*10^-6

now

pOH = -log(OH-) = -log( 2.58*10^-6) = 5.588

pH = 14-pOH = 14-5.588= 8.412

pH = 8.412

which is BASIC as expected

#### Earn Coins

Coins can be redeemed for fabulous gifts.