Question

How many grams of glycine amide (FM 74.08) should be used with 1.0 gram of glycine amide HCl (FM 110.54, pKa=8.20) to give 100 mL at pH 8.0?

The buffer equilibrium reaction is:
GlycineAmideHCl <--- ---> GlycineAmide + H+
which has the general form:
BH+ <--- Ka ---> B + H+


I'm assuming I need to use the henderson-hasselbalch equation which would give something like:

8 = 8.2 + log((B/BH+))

Though I'm not sure what the next step is or what to do with the B and BH+

Answer 1

Take you HH equation of
8.00 = 8.20 + log(B/A)
and solve for (B)/(A). I get about 0.63, then
base = 0.63A
You have 1 g of the acid and you know the molar mass, substitute for A which will allow you to calculate A, then solve for B, then determine how many grams that will be in 100 mL.

Answer 2

So 8.00 = 8.20 + log(B/A)
=> -0.2 = log(B/A)
=> 10^-0.2 = 10^log(B/A)
=> 0.63 = B/A
=> base = 0.63A as you said

Though I'm still having trouble understanding what you mean by "substitute for A which will allow you to calculate A, then solve for B"

Answer 3

Ohh, I got it now =)

Answer 4

good.