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.
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+
4 answers
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"
=> -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"
Ohh, I got it now =)
good.
1 answer