Question

the dissociation constant for acetic acid and HCN at 25 degree C are 1.5 * 10-5 and 4.5 * 10-10



respectively.The equilibrium constant for the equilibrium



CN- + CH3COOH -----------



3COO- would be what? Explain plzzz!!

Answer 1

The equilibrium constant for the equilibrium $CN^{-} +CH_{3} COOH$ and $HCN+CH_{3} COO^{-}$ is $3.3*10^{4}$

Explanation:

Consider the equation for equilibrium constant

$CH_{3} COOH$⇆$CH_{3} COO^{-} +H^{+}$

$k_{a} =1.5*10^{-5}$ for above equation

$HCN$⇆$H^{+} +CN^{-}$

Here $K_{a} =4.5*10^{-10}$ for above equation

Then

$CN^{-} +CH_{3} COOH$⇆$HCN+CH_{3} COO^{-}$

So we have to find the equilibrium constant K value for above equation

For determining the K value we have one equation

$K=\frac{K_{a} }{K_{a1} } =\frac{1.5*10^{-5} }{4.5*10^{-10} }$

         $=\frac{10^{5} }{3} =3.33*10^{4}$

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