# Preparing 1M acetic acid buffered with 1 M Ca-acetate to specific pH

I am prepping a buffer solution of 1M acetic acid and 1M calcium acetate with a pH of 5, but it has been a long time since chem 101. I have some notes from an old lab mate that suggested that 600mL of acetic acid and 150mL Ca-acetate (with water added to make 1L) makes about a pH of 4.5. However when try using Henderson-Hasselbalch to get amount for a pH of 5 I get vastly different results - 640mL Ca-acetate and 360 mL of acetic acid. Where am I going wrong? Any tips on how to calculate the amounts needed correctly?

Thank you

This is the way I would approach the problem, for what it's worth.

Henderson Hasselbalch equation is pH = pKa + log [base]/[acid]
If we take the pKa of acetic acid (HAc) to be 4.74, then we have
5 = 4.74 + log [B]/[A] where B is the base and A is the acid (B = acetate)
Solving for B/A we get
log [B]/[A] = 5 - 4.74 = 0.26
[B]/[A] = 1.82 and ...
[B] = 1.82[A]
But since you are using calcium acetate as the salt, there are two acetates for each calcium acetate, i.e. CaAc_2 ==>2Ac^- + Ca^2+
Thus, [B] = 0.91[A] and
A + B = 1000 ml (to make 1 liter of buffer)
Simultaneous equations and solve for A
0.91A + A = 1000 ml
1.91A = 1000 ml
A = 524 ml
B = 476 ml

• Sorry for poor formatting. I'm new here. – Dr. J. May 17 '18 at 17:16
• Ok, thanks. I was definitely missing the two acetate ions part. – RNocked May 18 '18 at 18:30