A25.0-ml sample of 0.150 m nitrous acid is titrated with a 0.150 m naoh solution. what is the ph after 13.3 ml of base is added? the ka of nitrous acid is 4.5 ⋅ 10-4.

8.11

Explanation:

Nitrous acid, HNO2 is a weak and monobasic acid. The equation of reaction is given below;

HNO2 + NaOH > NaNO2 + H2O

STEP 1: Find the number of NO^2- formed;

0.150 mol HNO2/L × 0.025L

=0.00375 mol of NO2^- formed.

STEP 2: calculate the concentration of NO2^-.

From the question we know that 25.0mL of NaOH reacts with 25.0 mL of Nitrous acid,HNO2 making a total of 50mL= 0.050L.

Neutralizing to the equivalence point, we have;

Concetration of NO2^- = 0.00375 mol÷ 0.050L

=0.075 M NO2^-

STEP 3: Calculate Kb .

from; kw/ka= Kb.

10-^14/4.50×10^-4

Kb= 2.22×10^-11.

We can see than kb is very small, therefore we use mass action expression to solve for 'x'. We have;

Kb=2.22×10^-11 = x^2/0.075-x

Approximately, x^2/0.075

STEP 4: calculating the pH,

X^2= [OH^-] = √0.075 kb.

Therefore, pOH = - log [OH^-] = 5.89

pH= 14- 5.89

pH= 8.11.