Posted by Sarah on Sunday, March 30, 2008 at 1:01am.
I don't find that problem. I don't have that edition of the book. The latest I have is the Sixth editon by Skoog, West, Holler. No Crouch.
Anyway, here is what you do.
Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
Look up the Kas for H3PO4. My book lists
k1 = 7.11 x 10^-3, k2 = 6.34 x 10^-8; k3 = 4.2 x 10^-13.
You want to use a ka close to the desired pH; therefore, we shall use k2.
H2PO4^- + OH^- ==> HPO4^- + H2O
k2 = 6.34 x 10^-8; pK2 = 7.20.
Substitute into the HH equation to find the ratio of base to acid. You want a buffered solution at pH = 7; therefore,
pH = pK2 + log[(base)/(acid)]
7.00 = 7.20 + log B/A
log B/A = -0.20
B/A = 0.631 or
(Base) = 0.631(Acid) or
(HPO4^=) = 0.631*(H2PO4^-)
Consider the acid H2PO4^- and the addition of OH^- to it.
H2PO4^- + OH^- ==> H2O + HPO4^=
We start with 0.2 mol H2PO4^- (from 0.2 M x 1 L from which we wish to prepare the buffer).
We add x mols OH^-, that will form x mols H2O, x mols(that's Base) of HPO4^=, and it will leave H2PO4^- (that's Acid) = 0.2 - x
Substitute these into the ratio we obtained above.
B = 0.631(A)
x = 0.631(0.2-x)
x = 0.0774 = (base) = (HPO4^=)
0.2-x = 0.2 - 0.0774 = 0.1226 = (acid) = (H2PO4^-).
You can prove this will give a pH = 7.00 by
pH = pk2 + log (base/acid).
pH = 7.20 + log(0.0774/0.1226) = 7.00
SO, to prepare the buffer, we use 1 L 0.2 M H3PO4, add enough NaOH to neutralize COMPLETELY the first hydrogen ion (0.2 M x 1 L = 0.2 mols. mols = g/molar mass or g = mols x molar mass = 0.2 mols x 40 g/mol NaOH = 8 g NaOH. That gets you 0.2 mol NaH2PO4 (the acid you want), then add 0.0774 mols more NaOH and that will be 0.0774 x 40 g/mol NaOH = 3.10 g NaOH for a total of 3.10 + 8.00 = 11.10 g NaOH. I hope this helps. Let me know if you have any questions. Check my thinking. Check my arithmetic.
Thank you so much! I was so confused, and now I am explaining how to do it to other people in my study group! I find the skoog book really hard to follow, so I couldn't work out even the formulae to use.
Thanks again, looking forward to a great mark in this assignment now.
Sarah
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