What is the volume of a solution of 0.5 M NaOH that must be added to adjust the pH
from 4 to 9 in 20 mL of a 100 mM solution of phosphoric acid?
Its been about 4 years since I've done this stuff. All I know is the pKa of phosphoric acid is 2.15 and I'm assuming In need to use the Henderson-Hassleback equation.
One problem I am having in helping you is that I can't figures how you started with 20 mL of 0.1M H3PO4 and it has a pH of 4. The pH of a 0.1 M H3PO4 soln is <2. The second problem is that you give a pk1 but not a pk2 (suggesting pk2 is not needed) BUT the pH of H3PO4 in which the first H is completely neutralized is sqrt(k1k2) or pH about 5 or so. If all of this is true, one must add enough base to go PAST the first equivalence point and stop somewhere just before the second equivalence point (pH about 9.8).
To find the volume of a solution of 0.5 M NaOH required to adjust the pH from 4 to 9, we need to use the Henderson-Hasselbalch equation and some additional information. The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffer solution.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
pH is the desired pH
pKa is the dissociation constant of the acid (given as 2.15 for phosphoric acid)
[A-] is the concentration of the conjugate base (in this case, Na2HPO4)
[HA] is the concentration of the acid (in this case, H3PO4 or phosphoric acid)
We are given the initial pH (4) and the concentration of the phosphoric acid solution (100 mM). Let's assume the final pH after adding NaOH is 9.
1. Calculate the ratio of [A-]/[HA] using the Henderson-Hasselbalch equation:
4 = 2.15 + log ([A-]/[HA])
log ([A-]/[HA]) = 4 - 2.15
log ([A-]/[HA]) = 1.85
2. Convert the desired ratio to actual concentrations:
[A-]/[HA] = 10^(1.85)
[A-]/[HA] = 73.98
3. Since the concentration of the acid is given as 100 mM, we can calculate the concentration of the conjugate base ([A-]):
[A-] = [HA] * ratio
[A-] = 100 mM * 73.98
[A-] = 7398 mM
4. The concentration of NaOH is 0.5 M. Now, we can calculate the volume of NaOH needed to achieve the desired concentration of the conjugate base (7398 mM):
Volume of NaOH = ([A-] * Volume of acid solution) / Concentration of NaOH
Volume of NaOH = (7398 mM * 20 mL) / 0.5 M
Volume of NaOH = 295.92 mL
Therefore, approximately 296 mL of a 0.5 M NaOH solution should be added to adjust the pH from 4 to 9 in a 20 mL 100 mM phosphoric acid solution.