Question

When a 0.031M aqueous solution of a certain acid is prepared, the acid is 0.89% dissociated. Calculate the acid dissociation constant Ka of the acid. Round your answer to 2 significant digits.

Answer 1

`Ka=2.5x10^(-6)`

Step-by-step explanation:

Hello,

In this case, by knowing that in terms of the change
`x` due to the reaction extent, the percent dissociation is:

`\% Dissociation:(x)/([acid]_0)`

Thus, we compute
`x` as:

`x=\% Dissociation*[acid]_0=0.89\%*0.031M=2.759x10^(-4)M`

With that
`x`, we could assume the law of mass action for a typical dissociation:

`Acid\rightleftharpoons H^++HA^-`

As:

`Ka=(x*x)/([acid]_0-x)=(2.759x10^(-4)M*2.759x10^(-4)M)/(0.031M-2.759x10^(-4)M) \\\\Ka=2.5x10^(-6)`

Best regards.