Calculate the pH of the following 0.0500 M H3AsO4 and 0.0500 M NaH2AsO4?
Use the Henderson-Hasselbalch equation.
To calculate the pH of a solution containing both a weak acid and its conjugate base, you can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
In this case, H3AsO4 is the weak acid (HA), and NaH2AsO4 is its conjugate base (A-). We need to find the pKa value for H3AsO4 first.
The pKa value can be determined using the Ka value and vice versa. Luckily, the Ka value for H3AsO4 can be given. Let's assume the Ka value for H3AsO4 is 5.0 x 10^(-3).
To calculate the pKa, take the negative logarithm of the Ka value:
pKa = -log(5.0 x 10^(-3))
Now that we have the pKa value, we can proceed with the Henderson-Hasselbalch equation.
Given:
[H3AsO4] = 0.0500 M
[NaH2AsO4] = 0.0500 M
pKa = -log(5.0 x 10^(-3))
First, we need to determine the concentrations of the acid and its conjugate base. As H3AsO4 will be partially ionized, we can assume that the initial concentration of the acid and its conjugate base will be the same as the given concentration.
[HA] = 0.0500 M
[A-] = 0.0500 M
Now, substitute these values into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
pH = -log(5.0 x 10^(-3)) + log (0.0500 / 0.0500)
Simplifying further:
pH = -log(5.0 x 10^(-3)) + log (1)
pH = -log(5.0 x 10^(-3))
Using a calculator, -log(5.0 x 10^(-3)) is approximately 2.3.
Therefore, the pH of the solution containing 0.0500 M H3AsO4 and 0.0500 M NaH2AsO4 is approximately 2.3.