Respuesta :
Answer: The pH of the resulting solution will be 3.001
Explanation:
Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
We are given:
Moles of NaOH = 0.0224 moles
Molarity of nitrous acid = 0.475 M
Molarity of sodium nitrite = 0.302 M
Volume of solution = 150 mL = 0.150 L (Conversion factor: 1 L = 1000 mL)
Putting values in equation 1, we get:
[tex]\text{Moles of nitrous acid}=(0.475mol/L\times 0.150L)=0.07125mol[/tex]
[tex]\text{Moles of sodium nitrite}=(0.302mol/L\times 0.150L)=0.0453mol[/tex]
The chemical equation for the reaction of nitrous acid and NaOH follows:
[tex]HNO_2+NaOH\rightleftharpoons NaNO_2+H_2O[/tex]
I: 0.07125 0.0224 0.0453
C: -0.0224 -0.0224 +0.0224
E: 0.04885 - 0.0677
The power of the acid dissociation constant is the negative logarithm of the acid dissociation constant. The equation used is:
[tex]pK_a=-\log K_a[/tex] ......(2)
We know:
[tex]K_a[/tex] for nitrous acid = [tex]7.2\times 10^{-4}[/tex]
Using equation 2:
[tex]pK_a=-\log (7.2\times 10^{-4})=3.143[/tex]
To calculate the pH of the acidic buffer, the equation for Henderson-Hasselbalch is used:
[tex]pH=pK_a+ \log \frac{\text{[conjugate base]}}{\text{[acid]}}[/tex] .......(3)
Given values:
[tex][NaNO_2]=\frac{0.0677}{0.150}[/tex]
[tex][HNO_2]=\frac{0.04885}{0.150}[/tex]
[tex]pK_a=3.143[/tex]
Putting values in equation 3. we get:
[tex]pH=3.143-\log \frac{(0.0677/0.150)}{(0.04885/0.150)}\\\\pH=3.143-0.142\\\\pH=3.001[/tex]
Hence, the pH of the resulting solution will be 3.001
