How many milliliters of 0.5M NaOH must be added to 200 mL of 0.1M NaH2PO4 to make a
buffer solution with a pH of 6.9?
To determine the number of milliliters of 0.5M NaOH needed to create a buffer solution with a pH of 6.9, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
In this case, NaH2PO4 acts as the weak acid (HA) and Na2HPO4 (formed by the reaction of NaOH with NaH2PO4) acts as the conjugate base (A-). The pKa of NaH2PO4 is determined by taking the negative logarithm of its acid dissociation constant (Ka).
Step 1: Calculate pKa
Since NaH2PO4 is a weak acid composed of one acidic hydrogen, the pKa can be determined using the Ka expression:
Ka = [H+][A-] / [HA]
Since NaH2PO4 completely dissociates in water, the concentration of the acid [HA] is equal to the initial concentration (0.1M). Since the pH of the buffer is 6.9, we can determine the concentration of H+ ions using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-6.9) = 1.2589 x 10^(-7) M
Now, we can rearrange the Ka equation to solve for [A-]:
Ka = [H+][A-] / [HA]
0.1 = (1.2589 x 10^(-7))(A-) / 0.1
Therefore,
A- = (0.1 x 0.1) / (1.2589 x 10^(-7))
A- = 0.000796 M
Now, we can determine the pKa using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
6.9 = pKa + log (0.000796 / 0.1)
pKa = 6.9 - log (0.000796 / 0.1)
pKa = 6.9 - 3.7
pKa = 3.2
Step 2: Calculate the moles of NaH2PO4
Since we have the initial concentration (0.1M) and volume (200 mL) of NaH2PO4, we can calculate the moles of NaH2PO4:
moles = concentration x volume
moles = 0.1 x (200/1000) L
moles = 0.02 mol
Step 3: Calculate the moles of NaOH required
Since 1 mole of NaH2PO4 reacts with 2 moles of NaOH to form Na2HPO4, we can determine the amount of NaOH we need:
moles of NaOH = 2 x moles of NaH2PO4
moles of NaOH = 2 x 0.02
moles of NaOH = 0.04 mol
Step 4: Calculate the volume of 0.5M NaOH
Finally, we can determine the volume of 0.5M NaOH needed using the molarity equation:
moles = concentration x volume
0.04 = 0.5 x (volume/1000) L
volume = (0.04 x 1000) / 0.5
volume = 80 mL
Therefore, you need to add 80 mL of 0.5M NaOH to 200 mL of 0.1M NaH2PO4 to create a buffer solution with a pH of 6.9.
200 mL x 0.1M H2PO4^- = 20 millimols.
?mL x 0.5M NaOH =?
........H2PO4^- + OH^- ==> HPO4^- + H2O
I.......20........0..........0........0
add...............x...................
C.......-x.......-x...........x
E.......20-x......0...........x
pH = pK2 + log(base)/(acid)
Substitute into HH equation the E line from the ICE chart above and solve for x = millimoles OH needed.
Then M = millimols/mL. You know millimoles and M, solve for mL.