Chemistry, asked by sam9876, 6 months ago

The pH of 0.001 M sodium butanoate is 9.0, then the dissociation constant Ka for butanoic acid will be at 298 K ???

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Answers

Answered by sarowashubham
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

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