Question

consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.

Answer 1

Answer:

$pH=10.5$

Explanation:

Hello,

In this case, the dissociation of the given weak acid is:

$HA\rightleftharpoons H^++A^-$

Therefore, the law of mass action for it turns out:

$Ka=\frac{[H^+][A^-]}{[HA]}$

That in terms of the change $x$ due to the reaction extent is:

$1x10^{-23}=\frac{x*x}{0.1-x}$

Thus, by solving with the quadratic equation or solver, we obtain:

$x=31.6x10^{-12}M$

Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:

$pH=-log(-31.6x10^{-12})\\pH=10.5$

Regards.