Answer these questions about glycine?

`a)`

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

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)`