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

`Ka=4.71x10^(-4)`

Step-by-step explanation:

Hello there!

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

`pH=pKa+log(([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(([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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