Buffer question?

Question

The amino acid glycine is often
used as the main ingredient of a buffer in biochemical experiments.
The amino group of glycine, which has a pKa of
9.6, can exist either in the protonated form (ONH3
) or as
the free base (ONH2), because of the reversible equilibrium
R-NH(3)+ ------>R-NH(2) + H+

(a) In what pH range can glycine be used as an effective
buffer due to its amino group?
(b) In a 0.1 M solution of glycine at pH 9.0, what fraction
of glycine has its amino group in the ONH3
form?
(c) How much 5 M KOH must be added to 1.0 L of 0.1 M
glycine at pH 9.0 to bring its pH to exactly 10.0?
(d) When 99% of the glycine is in its ONH3
form, what
is the numerical relation between the pH of the solution and
the pKa of the amino group?

How do i even start with this problem? Any hints appreciated.

Answer 1

This question looks familiar. Let's see if I can give a familiar answer:



"Well, here we go:

(a) Buffers are effective for pH = pKa +/- 1. Therefore, this is effective for pH of 8.6 to 10.6.

(b) This is almost purely rice, with a little algebra

Set y as the concentration of glycine already converted to free base.

R R-NH3+ <--> R-NH2 + H+
I | .1 - y | | y | 0
C | -x | | +x | | +x |
E | .1 -x -y | | y+x | | x |

Then you know that x is equal to [H+] which is 10^-9, or 1E-9.
You also know that Ka is 10^-9.6, which is 2.51E-10.

Algebraically, you can solve for y to find that if (y+1E-9)(1E-9)/(.1-y-1E-9) = 2.51E-10
then y = .0200639 M

That is the concentration of glycine as free base, leaving .0799361 M left as the protonated form.

.0799M/.1M is about 4/5. That is your answer.

(c) This involves the prior information plus a little more

We already know the concentrations before the NaOH is added. Now we have to find how much more NH3+ form we lose to reach pH = 10.0.

Set y as TOTAL concentration converted to NH2 (it's easier). (This will look really familiar.)

R R-NH3+ <--> R-NH2 + H+
I | .1 - y | | y | 0
C | -x | | +x | | +x |
E | .1 -x -y | | y+x | | x |

Then you know that x is equal to [H+] which is 10^-10, or 1E-10.
You also know that Ka is 10^-9.6, which is 2.51E-10.

Algebraically, you can solve for y to find that if (y+1E-10)(1E-10)/(.1-y-1E-10) = 2.51E-10
then y = .07151 M.

So the change in concentration of the free base is .07151 M - .02006 M = .05145 M

1.0 L * .05145 M = .05145 mol glycine converted

Since the glycine loses one proton to every one NaOH molecule, you know that .05145 mol NaOH were added.

.05145 mol / (5 mol/L) = .0102 L = 10.2 mL of NaOH.
That is the answer.

(d) This part is actually easier than (c) and (d). It is only RICE.

R R-NH3+ <--> R-NH2 + H+
I |.099| .001 0
C | -x | | +x | | +x |
E |.099 -x| | .001+x | | x |

(.001+x)(x)/(.099-x) = 2.51E-10

This makes x (conc. H+) be equal to 2.48E-8, so final pH is 7.61. This means that pH = pKa - 1.99. That is your answer.

OR

Much easier, of course, is Henderson-Hasselbalch:

pH = pKa + log([HA]/[A-]) = pKa + log(.99/.01) = pKa - 2. That is also your answer.

Hope that helps. Have a good day."

Answer 2

PH=11

Answer 3

a) 6.5 to 9.1

b) 47 %

c) 320 ml

d) Equal.