Respuesta :
Answer: 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.
The pH after 13.3 mL of base is added 8.11
Balanced chemical reaction:
HNO₂ + NaOH ---------> NaNO₂ + H₂O
Find the number of [tex]NO_2^-[/tex] formed:
= 0.150 mol HNO₂ /L × 0.025L
= 0.00375 mol of [tex]NO_2^-[/tex] formed.
Calculate the concentration of [tex]NO_2^-[/tex] :
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;
Concentration of [tex]NO_2^-[/tex] = 0.00375 mol / 0.050L
Concentration of [tex]NO_2^-[/tex] [tex]=0.075 M NO_2^-[/tex]
Calculate Kb:
From; kw/ka= Kb
[tex]10^{-14}/4.50*10^{-4}\\\\K_b= 2.22*10^{-11}[/tex]
We can see than kb is very small, therefore we use mass action expression to solve for 'x'. We have;
[tex]K_b=2.22*10^{-11}\\\\ K_b= x^2/0.075-x\\\\K_b= x^2/0.075[/tex]
Calculating the pH:
[tex]x^2= [OH^-] = \sqrt{0.075 K_b}[/tex]
Therefore, pOH = - log [OH-] = 5.89
pH= 14- 5.89
pH= 8.11
Find more information about pH here:
brainly.com/question/22390063
