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

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.

Answer 2

Answer:

$Ka=1.53x10^-5$

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:

$Ka=\frac{X*X}{0.3-X} =\frac{X^2}{0.3-X}$

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:

$Ka=\frac{(0.00213)^2}{0.3-0.00213}$

$Ka=1.53x10^-5$