The pH of 0.001 M sodium butanoate is 9.0, then the dissociation constant Ka for butanoic acid will be at 298 K ???
No Spam....
Answers
Answered by
2
Answer:
10^-6
Explanation:
I am showing butanoate ion by Bo
The reaction will be in water be like
BoNa + H20---------> BoH + NaoH
BoH is butanoic acid.
Now ph is 9 that mean ph =9
-log [H+] = 9
[H+] = 10^-9
dissociation constant
= conc. of product butanoic acid ÷ conc. of reactant sodium butanoate (0.001M)
conc. of butanoic acid can be taken nearly same to H+
conc.
so we get K= 10^-9 ÷ 10^-3 = 10^-6
Similar questions