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
Glycine is the simplest amino acid, containing two
"H" atoms on itsalpha carbon:
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The carboxyl proton
"pKa" is near2.2 , and the amine proton"pKa" is around9.8 . 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.
The
"pH" of"0.1 M" solutions of each of these forms is based on theK_a values. The most acidic form is"AA"^+ and the most basic form is"AA"^- .With
K_(a1) = 4.57 xx 10^(-3) andK_(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)