Question

The pH of 0.259 M solution of an unknown acid is measured to be 2.353. What is the Ka of the acid?

Answer 1

Answer:

$Ka=7.73x10^{-5}$

Explanation:

Hello!

In this case, since the pH is measure of the amount of hydrogen ions in solution, first given the pH we can compute the concentration of hydrogen ions via:

$pH=-log([H^+])$

$[H^+]=10^{-pH}=10^{-2.353}=4.436x10^{-3}M$

Thus, since the ionization of the given acid, leads to the following equilibrium expression as it is a weak one:

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

We notice the concentration of hydrogen ions equal the concentration of the acids conjugate base, so we can compute:

$Ka=\frac{4.436x10^{-3}*4.436x10^{-3}}{0.259-4.436x10^{-3}}\\\\Ka=7.73x10^{-5}$

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