Question

The ph of a 0.30 m solution of a weak acid is 2.67. what is the ka for this acid?

Answer 1

`Ka=1.53x10^-5`

Step-by-step explanation:

Since it is a weak acid it is necessary to write the balance. Where the unknown acid will be HA.

`HA<->\hspace{0.1cm}A^- \hspace{0.1cm}+\hspace{0.1cm}H^+`.

Then with the equilibrium we can write the ICE table (figure 1).

Where we obtain the equation:

`Ka=(X*X)/(0.3-X) =(X^2)/(0.3-X)`

Now, we have to find "X". To do this we can use the pH equation and the pH value:

`pH=-Log\hspace{0.1cm}H^+`

`-pH=Log\hspace{0.1cm}H^+`

`10^-pH=H^+`

`10^-2.67=H^+`

`H^+=0.00213`

`X=0.00213`

Then we have to plug the values in the Ka equation:

`Ka=((0.00213)^2)/(0.3-0.00213)`

`Ka=1.53x10^-5`

Answer 2

To determine the Ka of the acid, we can use the equation for the pH of weak acids which is expressed as:

pH = -0.5 log Ka
2.67 = -0.5 log Ka
Ka = 4.571x10^-6

Weak acids are acids that do not dissociate completely in solution. The solution would contain the cations, anions and the acid itself as a compound. Hope this helps.