# Dissociation of glycine

1. Sep 16, 2005

### 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

2. Sep 17, 2005

### GCT

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.

3. Sep 17, 2005

### apchemstudent

4. Sep 17, 2005

### GCT

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.