Question

During a titration, the pH of an analyte solution containing HA(aq) is 3.92 and the ratio of [A–]/[HA] is 0.41. What is the Ka of HA?

Answer 1

Answer:

$Ka=4.71x10^{-4}$

Explanation:

Hello there!

In this case, according to the Henderson-Hasselbach equation, it is possible to write:

$pH=pKa+log(\frac{[A^-]}{[HA]} )$

Next, since we are given the pH and the [A–]/[HA] ratio, we can solve for the pKa as shown below:

$pKa=pH-log(\frac{[A^-]}{[HA]} )$

Now, we plug in the values to obtain:

$pKa=3.92-log(0.41 )\\\\pKa=3.33$

Next, Ka is:

$Ka=10^{-pKa}=10^{-3.33}\\\\Ka=4.71x10^{-4}$

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