Answer these questions about glycine?

a) Describe the structure and charge(s) of glycine below, at, and above "pH" 7.4.
b) What is the "pH" of each form of glycine in separate aqueous solutions at body temperature, if the concentration is "0.1 M" for them initially? K_(a1) = 4.57 xx 10^(-3), K_(a2) = 2.51 xx 10^(-10).

1 Answer
Nov 11, 2017

a)

Glycine is the simplest amino acid, containing two "H" atoms on its alpha carbon:

http://2.bp.blogspot.com/

The carboxyl proton "pKa" is near 2.2, and the amine proton "pKa" is around 9.8. Thus,

  • Below "pH" ~~ 2.2, the amine proton is ON, and the carboxyl proton is ON. This is the "AA"^+ form.
  • At "pH" = 7.4 (body "pH"), the amine proton is ON, and the carboxyl proton is OFF. This is the "AA"^0 form.
  • Above "pH" ~~ 9.8, the amine proton is OFF, and the carboxyl proton is OFF. This is the "AA"^- form.

You can draw those by interpreting the notes above.

b)

The "pH" of "0.1 M" solutions of each of these forms is based on the K_a values. The most acidic form is "AA"^+ and the most basic form is "AA"^-.

With K_(a1) = 4.57 xx 10^(-3) and K_(a2) = 2.51 xx 10^(-10), the first one can be seen to belong to the carboxyl proton (why? It's higher, so it is ascribed to the more acidic proton). Treating glycine as a diprotic acid,

"H"_2"Gly"^(+)(aq) rightleftharpoons "HGly"(aq) + "H"^(+)(aq)

"I"" "" ""0.1 M"" "" "" "" ""0 M"" "" "" "" ""0 M"
"C"" "" "-x" "" "" "" "+x" "" "" "" "+x
"E"" "(0.1-x)"M"" "" "" "x" M"" "" "" "x" M"

And so, write the mass action expression:

K_(a1) = 4.57 xx 10^(-3) = x^2/(0.1 - x)

Naively using the small x approximation,

4.57 xx 10^(-3) ~~ x^2/0.1

=> x_0 ~~ sqrt(4.57 xx 10^(-3) cdot 0.1) = "0.0214 M"

That is not yet correct. We then generate our next guesses to converge onto the final x:

=> x_1 ~~ sqrt(4.57 xx 10^(-3) cdot (0.1 - 0.0214)) = "0.0190 M"
=> x_2 ~~ sqrt(4.57 xx 10^(-3) cdot (0.1 - 0.0190)) = "0.0192 M"
=> x_3 ~~ sqrt(4.57 xx 10^(-3) cdot (0.1 - 0.0192)) = "0.0192 M"

And so, ["H"^(+)] = "0.0192 M", giving

color(blue)("pH") = -log(0.0192) = color(blue)(1.72).

The first conjugate base is a much weaker acid. Again, we treat this as in a separate solution:

"HGly"(aq) rightleftharpoons "Gly"^(-)(aq) + "H"^(+)(aq)

"I"" "" ""0.1 M"" "" "" ""0 M"" "" "" ""0 M"
"C"" "" "-x" "" "" "+x" "" "" "+x
"E"" "(0.1-x)"M"" "" "x" M"" "" "x" M"

And so, write the mass action expression:

K_(a2) = 2.51 xx 10^(-10) = x^2/(0.1 - x) ~~ x^2/0.1

Here, the small x approximation is justified; K_(a2) "<<" 10^(-5). Thus:

x = ["H"^(+)] = sqrt(2.51 xx 10^(-10) cdot 0.1) = 5.01 xx 10^(-6) "M"

And so,

color(blue)("pH") = -log["H"^(+)] = -log(5.01 xx 10^(-6)) = color(blue)(5.30).

Its conjugate base is obviously a base, and so, treating this as in ANOTHER separate solution, we need the K_b.

At body temperature, 37^@ "C", K_w = 2.42 xx 10^(-14), so:

K_b = K_w/K_a = (2.42 xx 10^(-14))/(2.51 xx 10^(-10)) = 9.64 xx 10^(-5)

This is still small enough for the small x approximation.

"Gly"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HGly"(aq) + "OH"^(-)(aq)

"I"" "" ""0.1 M"" "" "" "-" "" "" ""0 M"" "" "" ""0 M"
"C"" "" "-x" "" "" "" "-" "" "+x" "" "" "+x
"E"" "(0.1-x)"M"" "" "-" "" "" "x" M"" "" "x" M"

Just as before...

K_b = 9.64 xx 10^(-5) = x^2/(0.1 - x) ~~ x^2/0.1

Thus, since ["OH"^(-)] = sqrt(K_b["Gly"^(-)]), the ["H"^(+)] is:

["H"^(+)] = K_w/(["OH"^(-)]) = (2.42 xx 10^(-14))/(sqrt(9.64 xx 10^(-5) cdot 0.1))

= 7.79 xx 10^(-12) "M"

This gives a "pH" of

color(blue)("pH") = -log(7.79 xx 10^(-12)) = color(blue)(11.1)