Wednesday
April 23, 2014

Homework Help: Chemistry

Posted by James on Tuesday, May 1, 2007 at 8:26pm.

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 know that [H+] is very nearly equal to Ka, and that pKa is very nearly equal to pH, so I figured that if I could find [H+] I could find pH for at least part a. My problem right now is that I'm having trouble writing out equations for this reaction. Do I treat each part as its own separate entity, so that I have
KH2PO4 --> K+ + H2PO4-
K2HPO4 --> 2K+ + HPO4-2
? Or do I have a reaction directly between KH2PO4 and K2HPO4, or neither? Also, how does the addition of HCl and NaOH affect the pH calculationwise? I figured that upon the addition of HCl and NaOH, water and neutral NaCl would be formed, which it seems like it would keep the pH constant after the addition of NaOH. But I'm still a little confused on this- if anyone could clarify this for me, that would be great.

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.

b) When x mols HCl are added, x mols HCl reacts with x mols of the base which add x mols to the acid and subtracts x mols from the base. Plug those numbers into a new HH equation, using the same pK2 and calculate the new pH.

c) When x mols NaOH are added, x mols NaOH reacts with x mols of the acid and that adds x mols to the base and subtracts x mols from the acid. Plug those new numbers into the HH equation and recalculate pH using the same pK2.

Post your work if you get stuck.

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