# Dissociation of glycine

apchemstudent
Here's the problem how much of 5M KOH should be added to 1L of 0.1M glycine(pKa=9.6) at a pH of 9 until the pH reaches exactly 10? the answer in the back says 10mL.

R-NH3(+) <-> R-NH2 + H(+)

Here's the way I solved it. I figured out the number of moles of R-NH2 using the henderson-hasselbach equation, and it is .251 when the pH was 9.

And then i figured out the increase in moles of R-NH2 when the pH is at 10:

pH = pKa + log(NH2+x / NH3(+) - x)
10 = 9.6 + log((moles of NH2+x)/(moles of NH3(+) - x))

x = .064

then for every mole of R-NH2 dissociated a mole of KOH must be used.

so concentration * volume = .064
5M * V = .064
V ~ 13 mL

This is not the correct answer, although it might seem close. How did they end up with 10 mL? I tried rounding it to 0.06 but that only gave me 12 mL. Am i being too precise here, and the book just rounded down to 10 mL due to significant figures? Please help. Thanks

Homework Helper
You're solution seems fine, if your not convinced try working backwards, you'll see that the the 13mL corresponds to the pH of 10.

apchemstudent

Homework Helper
ahh I get it, the .1M glycine doesn't represent the acid, it represents glycine, the molecule, as a whole. The nomenclature's all **** up. Well, there's biochemistry for ya, nomenclature in genetics is also a travesty. Unfortunately I don't believe they have the name for the different protonated forms of glycine.

the .251 is the ratio of the base to the acid, to find the concentration of the base

$$(.251/1.251).1~=~initial~base~conc.$$
$$(.1/1.251)=~initial~acid~conc.$$

$$K=[.02006+x]10^{-10}/(.07994-x)=2.51x10^{-10}$$

$$x=.05145=v5/(v+1)$$

$$v=~10mL$$

If anyone needs any clearing up, just say so, I admit the solution's pretty messy.