# Chemistry

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I don't understand how to work out the volumes of weak acid and strong base to mix to make 1L of buffer when given the concentrations of weak acid and strong base (in my case 0.160M NaOH and 0.200M H3PO4) and the pH required (7.00).

Specifically the textbook is Fundamentals of Analytical chem by skoog. west, holler, and crouch, and the question is 15-21.
Thanks

• ChemiDrBob222 - ,

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.

• ChemiDrBob222 - ,

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