I posted a question earlier that I was having trouble with, but I was unable to attempt to try it again for several days after posting. The question was:
A buffer is made up of .300 L each of .500 M KH2PO4 and .317 M K2HPO4. Assuming tha the volumes are additive, calculate
a) the pH of the buffer
b) the pH of the bugger after the addition of .0500 mol of HCl to .600 L of the buffer.
c) the pH of the buffer after the addition of .0500 mol of NaOH to .600 L of the buffer.
I got the following response for A (which is what I'm struggling with at the moment):
a) With a mixture of KH2PO4 and K2HPO4, you are between the 1st and 2nd equivalence points in the titration of H3PO4 with KOH. To calculate the pH, use the Henderson-Hasselbalch equation.
pH = pKa + log (base)/(acid)
H3PO4 has k1, k2, and k3. For this problem, pKa is pK2. That makes the HPO4^-2 the base and H2PO4^- the acid. Straight forward HH equation.
What I'm really having trouble with is setting up the equations so that I can find the concentrations. What I have written down is this, but I can't figure out if it's correct or not:
KH2PO4---> K+ + H2PO4-
K2HPO4---> 2K+ + HPO4-2
where H2PO4- is the acid, and H2PO4-2 is the base.
I then have the Ka of H2PO4- as 6.2 x 10^-8 (this is the value that I found in the book. I continued the problem like so...:
pH= (-log(6.2x10^-8)) + log .150/.0950= 7.506
I'm not sure if the concentrations are correct. Does this seem right so far?
Also, I'm trying to set up equations for b and c... do these look correct?
b) HPO4-2 + HCl --> H2PO4- + Cl-
c) H2PO4- + NaOH --> HPO4-2 + H2O + Na+
?
The definition of molarity is all you need for the concentration. Molarity = #mols/L.
(KH2PO4) = (0.3L x 0.15 M)/0.6L
(H2HPO4)= (0.3L x 0.317 M)/0.6L
pH = pK2 + log (b/a)
To calculate the pH of the buffer solution, you're on the right track with the Henderson-Hasselbalch equation.
The equations you wrote for KH2PO4 and K2HPO4 are correct.
To find the concentrations of KH2PO4 and K2HPO4, you can use the formula for molarity: Molarity (M) = moles of solute / volume of solution in liters.
For KH2PO4:
Molarity = (0.15 mol / 0.6 L) = 0.25 M
For K2HPO4:
Molarity = (0.317 mol / 0.6 L) = 0.5283 M
Now you can substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log (base)/(acid)
Since you are dealing with the second ionization constant of phosphoric acid (H3PO4), which is pK2, the base is HPO4^-2 and the acid is H2PO4^-.
pH = pK2 + log [(0.5283 M) / (0.25 M)]
Now, you mentioned that you found the pK2 value to be 6.2 x 10^-8 in the book. If this is correct, you can substitute it into the equation and calculate the pH:
pH = (-log(6.2 x 10^-8)) + log [(0.5283 M) / (0.25 M)]
Calculate the logarithms and perform the addition to find the pH.
Regarding parts b) and c), your equations are correct for the addition of HCl and NaOH, respectively. Keep in mind that HCl will contribute additional H+ ions to the solution, while NaOH will contribute additional OH- ions. You can use the Henderson-Hasselbalch equation again to calculate the new pH values using the updated concentrations after the addition of these compounds.
I hope this helps! Let me know if you have any further questions.