Question

Glycine `[NH_(2)CH_(2)COOH)` is basic and acidic due to presence of `-NH_(2)` and `-COOH` group. It acquires a `H^(o+)` to form `overset(o+)NH_(3)COOH`, which is a diprotic acid with `K_(1) = 4.5 5 10^(-3)` and `K_(2) = 1.7 xx 10^(-10)`. In a `0.01M` solution of neutral glycine,
a. What is the `pH` and
b. What percent of the glycine is in the cationic form at equilibrium?

Answer 1

a. Glycine is represented as `HG1Y`:
`(I)H_(2)G1Y^(o+) H_(2)O hArr Hgly + H_(3)O^(o+)`
`K_(1) = ([HG1Y][H_(3)O^(o+)])/([H_(2)G1Y^(o+)])=K_(w) = [H_(3)O^(o+)][overset(Theta)OH]`
`K_(h) = (K_(w))/(K_(1)) = ([cancel[H_(3)O^(o+))][overset(Theta)OH][H_(2)G1y^(o+)])/([HG1y][cancel[H_(3)O^(o+))])`
`= ([H_(2)G1y^(o+)][overset(Theta)OH])/([HG1y])`
`K_(h) = (10^(-14))/(4.5xx10^(-3)) = 2.2 xx 10^(-12)`
`(II) HG1y + H_(2)O hArr G1y^(Theta) = H_(3)O^(o+)`
`K_(2) = ([G1y^(Theta)][H_(3)O^(o+)])/([HG1y])=1.7 xx 10^(-10)`
Assuming `[H_(2)G1y^(o+)] = [G1y^(Theta)]`
`(K_(2))/(K_(h)) = ([G1y^(Theta)][H_(3)O^(o+)][cancel[HG1y)])/([cancelHG1Y]xx[H_(2)G1y^(o+)][overset(Theta)OH]) =([H_(3)O^(o+)])/([overset(Theta)OH])`
`(1.7 xx 10^(-10))/(2.2 xx 10^(-12)) = 75 = ([H_(3)O^(o+)])/([overset(Theta)OH])`
Multiply equation (i) by `K_(w)` equation.
`([H_(3)O^(o+)])/([cancel[overset(Theta)OH)])xx[H_(3)O^(o+)][cancel[overset(Theta)OH)] = [H_(3)O^(o+)] = 75xx10^(-14)`.
`:. [H_(3)O^(o+)] = 8.7 xx 10^(-7)` and `pH = 6.06`
b. `K_(1) = ([HG1y][H_(3)O^(o+)])/([H_(2)G1y^(o+)])`
`4.5 xx 10^(-3) = ((0.01)(8.7 xx 10^(-7)))/([H_(2)G1y^(o+)])`
`:. [H_(2)G1y^(o+)] = 1.9 xx 10^(-6) M`
`%` cationic form `= (1.9 xx 10^(-6))/(0.01) xx100 =0.019%`