You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-
ID: 948731 • Letter: Y
Question
You prepare a buffer solution from 10.0 mL of 0.100 M MOPS (3-morpholinopropane-1-sulfonic acid) and 10.0 mL of 0.079 M NaOH. Next, you add 1.00 mL of 5.67 × 10^-5 M lidocaine to this mixture. Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH+.
MOPS Ka = 6.3 × 10^–8.
Lidocaine Kb = 8.7 × 10^–7
(Hint: First calculate the pH of the solution from the amount of MOPS added. Start by finding the number of moles (or millimoles) of HA and OH–. Which reactant is limiting? Next, find the amount of A– that forms and the amount of HA left over. Finally, determine the [A–]/[HA] ratio and use the Henderson-Haselbalch equation to find pH. The concentration of lidocaine is too low to affect this pH value. Calculate the pKa of lidocaine and then the fraction in the protonated form from the pH and the pKa using the Henderson Hasselbalch equation.)
Explanation / Answer
We have Ka of MOPS = 6.3 × 10^–8
Kb of Lidocaine = 8.7 × 10^–7
First we need to calculate pH of the solution
10 mL * 0.1 MOPS acid = 1.0 mm
10 mL * 0.079 = 0.79 mm NaOH
MOPSacid + OH- ==> MOPSbase + H2O
I......1..........0.........0
C......-0.79...-0.79........0.79
E.......0.21......0.79.........0.79
pKa = -log10(Ka) =-log10 (6.3 × 10^–8) = 6.200
pH = pKa + log10([A-]/[HA+]
= 6.2 + log10([0.79]/[0.21]
= 6.2 + 0.575 = 6.775
[H+] = 10^-pH = 10^(-6.775) = 1.678 * 10^-7
The concentration of lidocaine is too low to affect given pH value.
Kb of lidocaine is 8.7 × 10^–7
We need to calculate pKb of lidocaine
pKb = -log10(Kb) =-log10 (8.7 × 10^–7) = 5.060
pKa = 14 – pKb = 14 – 5.060 = 8.94
Now, we will use the Henderson Hasselbalch equation to calculate LH+
We will be using same value of [H+]
pH = pKa + log10([A-]/[H+])
6.775 = 8.94 + log10([A-]/[H+])
log10([A-]/[H+]) = 6.775 - 8.94 = -2.165
([A-]/[H+]) = antilog (-2.165) = 0.0068
But we need to calculate fraction of [H+]
[H+] = [A-]/0.0068 = 0.79 / 0.0068
= 116.17
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