January 19, 2017

Homework Help: chem

Posted by James on Saturday, May 5, 2007 at 8:25pm.

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)

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