How to estimate pka from ph?

The following formula is used to calculate the pKa.

Variables:

To calculate pKa, subtract the value of the log of the conjugate base concentration divide by the weak acid concentration from the pH value.

The following steps outline how to calculate the pKa.

Example Problem:

Use the following variables as an example problem to test your knowledge.

pH value = 5

concentration of conjugate base = .32

concentration of weak acid = .45

• First, determine the pH value.
• Next, determine the concentration of conjugate base.
• Next, determine the concentration of weak acid.
• Next, gather the formula from above = pKa = pH – Log(CB/WA).
• Finally, calculate the pKa.

The relationship between pH and pKa is described by the Henderson-Hasselbalch equation.

Once you have pH or pKa values, you know certain things about a solution and how it compares with other solutions:

If you know either pH or pKa, you can solve for the other value using an approximation called the Henderson-Hasselbalch equation:

pH = pKa + log ([conjugate base]/[weak acid])pH = pka+log ([A-]/[HA])

pH is the sum of the pKa value and the log of the concentration of the conjugate base divided by the concentration of the weak acid.

At half the equivalence point:

pH = pKa

It's worth noting sometimes this equation is written for the Ka value rather than pKa, so you should know the relationship:

pKa = -logKa

The reason the Henderson-Hasselbalch equation is an approximation is because it takes water chemistry out of the equation. This works when water is the solvent and is present in a very large proportion to the [H+] and acid/conjugate base. You shouldn't try to apply the approximation for concentrated solutions. Use the approximation only when the following conditions are met:

Find [H+] for a solution of 0.225 M NaNO2 and 1.0 M HNO2. The Ka value (from a table) of HNO2 is 5.6 x 10-4.

pKa = −log Ka = −log(7.4×10−4) = 3.14

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

pH = pKa + log([NO2-]/[HNO2])

pH = 3.14 + log(1/0.225)

pH = 3.14 + 0.648 = 3.788

[H+] = 10−pH = 10−3.788 = 1.6×10−4