the dissociation equilibrium constant for the protonated form The dissociation e
ID: 527270 • Letter: T
Question
the dissociation equilibrium constant for the protonated form The dissociation equilibrium constants for the protonated form of alanine (a diprotic amino acid, H_2X^+) are K_al = 4.6 times 10^-3 and K_a2 = 2.0 times 10^-10. Assume that you start with 50.00 mL of a 0.050 M solution of alanine. What is the pH of this initial solution? () What is the concentration of HX after 25.0 mL of 0.100 M NaOH has been added to the solution? () What is the pH of the solution after 25.0 mL of 0.100 M NaOH has been added to the solution? ()Explanation / Answer
50.00 mL of a 0.0500 M solution of alanine is: 0.05000*0.0500 = 0.00250 mol of alanine.
25.00 mL of 0.100 M NaOH is: 0.02500*0.100 = 0.00250 mol of NaOH.
0.00250 mol of alanine reacts with 0.00250 mol of NaOH to form 0.00250 mol of mono-protic single-charged ion in 75.00ml solution (assume the volume does not change after mixing). The concentration is: [A-] = 0.00250/0.07500 M = (1/30) M.
Let [A] represent neutral alanine concentration, [A-] the mono-protic single-charged ion, and [A--] the double-charged ion. Also let us assume that in the solution at equilibrium, X M [A--] and Y M [A] will be formed at the cost of [A-]. We have in formula:
. . . . . . . . . . . . . A A- . . + . . H+ . . A(2-) + 2H+
Initially: . . . . . . .0 . . . 1/30 . . . trace . . . .0 . . . . trace
At equilibrium: Y . .1/30-X-Y . . X-Y . . . . X . . . . .X-Y
and we have from the definition of dissociation equilibrium constants:
K1 = 0.0046 = [A-]*[H+]/[A] = (1/30)*(X-Y)/Y . . . . . . .(1)
K2 = 2.0x10^-10 = [A--]*[H+]/[A-] = X*(X-Y)/(1/30) . .(2)
where we already take the approximation:
1/30 -X -Y 1/30. You will see that the approximation is valid and it would save us quite a big trouble of math manipulation.
From (1), we have: 0.0046*30*Y = X-Y
or: 2.38Y = X . . . . . (3)
Put (3) into (2), we have: 2.38Y*(1.38Y)*30 = 2.0x10^-10
or: Y = 1.42x10^-6 (M)
Now you see that 1/30 >> X+Y
Thus [H+] = X-Y = 1.38Y = 1.97x10^-6 (M)
pH = -log([H+]) = -log(1.97x10^-6) = 5.71
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