115k views
0 votes
A 300 ml 1.6 M solution of an acid (HA), has a pH of

2.4 at 25oC. What is the Ka for this acid?? Using the Henderson-
Hass. Equation, calculate the A- (conjugate base Molarity) you need to
get a pH of 4.6.

1 Answer

0 votes

To find the Ka for the acid, we can use the Henderson-Hasselbalch equation and the given pH value. Using the Henderson-Hasselbalch equation again, we can calculate the concentration of [A-] required to achieve a specific pH. By substituting the known values and solving the equations, we can determine the Ka for the acid and the concentration of [A-] needed for a pH of 4.6.

To find the Ka for the acid, we can start by using the pH value and the Henderson-Hasselbalch equation:

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

In this equation, pKa represents the negative logarithm of the acid dissociation constant (Ka). [A-] and [HA] represent the concentrations of the conjugate base and the acid, respectively.

Given that the pH is 2.4, we can rearrange the equation to solve for pKa:

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

Substituting the values, we have:

pKa = 2.4 - log([A-]/[HA])

Next, we need to calculate the concentration of [A-] required to achieve a pH of 4.6. We can rearrange the Henderson-Hasselbalch equation once again to solve for [A-]:

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

Substituting the values, we have:

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

To isolate [A-], we can rearrange the equation:

log([A-]/[HA]) = 4.6 - pKa

Next, we can calculate the antilog of both sides of the equation to solve for [A-]:

[A-]/[HA] = 10^(4.6 - pKa)

Finally, we can calculate the concentration of [A-] using the given values and the calculated pKa:

[A-] = [HA] * 10^(4.6 - pKa)