Answer :
Answer:
[tex]Ka=3.98x10^{-4}[/tex]
Explanation:
Hello there!
In this case, since the modelling of titration problems can be approached via the Henderson-Hasselbach equation to set up a relationship between pH, pKa and the concentration of the acid and its conjugate base, we can write:
[tex]pH=pKa+log(\frac{[NO_2^-]}{[HNO_2]} )[/tex]
Whereas the pH is given as 3.14 and the concentrations are the same, that is why the pH would be equal to the pKa as the logarithm gets 0 (log(1)=0); thus, we can calculate the Ka via:
[tex]Ka=10^{-pKa}=10^{-3.14}\\\\Ka=3.98x10^{-4}[/tex]
Best regards!