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

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.

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jescaherga jescaherga · Answer 2 · 2019-10-21T19:12:29+02:00

Answer:

[tex]Ka=1.53x10^-5[/tex]

Explanation:

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

[tex]HA<->\hspace{0.1cm}A^- \hspace{0.1cm}+\hspace{0.1cm}H^+[/tex].

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

Where we obtain the equation:

[tex]Ka=\frac{X*X}{0.3-X} =\frac{X^2}{0.3-X}[/tex]

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

[tex]pH=-Log\hspace{0.1cm}H^+[/tex]

[tex]-pH=Log\hspace{0.1cm}H^+[/tex]

[tex]10^-pH=H^+[/tex]

[tex]10^-2.67=H^+[/tex]

[tex]H^+=0.00213[/tex]

[tex]X=0.00213[/tex]

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

[tex]Ka=\frac{(0.00213)^2}{0.3-0.00213}[/tex]

[tex]Ka=1.53x10^-5[/tex]