What is the pH of a 0.031 mol L-1 solution of an acid, if the pKa of the acid is 5.12?

To solve this problem, we can use the Henderson-Hasselbalch equation, which is:

pH = pKa + log([A-]/[HA])

In this case, we are dealing with a weak acid that is not in a buffer solution, so the concentration of the acid ([HA]) is equal to the initial concentration of the acid, and the concentration of the conjugate base ([A-]) is negligible. Therefore, the equation simplifies to:

pH = pKa + log(0)

The log of 0 is undefined, so we can't use this equation directly. However, we can make an approximation. If the concentration of the acid is much less than 10^-pKa, then the pH is approximately equal to the pKa. In this case, the concentration of the acid is 0.031 mol L-1, and 10^-5.12 is approximately 0.0000076 mol L-1. Since 0.031 is much greater than 0.0000076, we can't make this approximation.

Instead, we need to use the definition of pH and pKa:

pH = -log[H+] pKa = -log(Ka)

We can solve the second equation for Ka:

Ka = 10^-pKa

Substituting the given value of pKa:

Ka = 10^-5.12

Next, we can use the definition of Ka for a weak acid HA:

Ka = [H+][A-]/[HA]

Since [A-] is negligible, this simplifies to:

Ka = [H+]/[HA]

We can solve this equation for [H+]:

[H+] = Ka[HA]

Substituting the values we found for Ka and [HA]:

[H+] = 10^-5.12 * 0.031

Finally, we can substitute this into the equation for pH:

pH = -log(10^-5.12 * 0.031)

This will give you the pH of the solution.