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 0.084 M NaOH. Next, you add 1.00 mL of 4.69 × 10-6 M lidocaine to this mixture Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH .
The Ka of MOPS is 6.3 e-8, and the Kb of lidocaine is 8.7e-7

Answer 1

moles of MOPS(HA)=0.1 mol/L*10.0ml= 0.1 mol/L*10ml 10^-3 L/ml=10^-3 moles

moles of NaOH=0.084 mol/L*10.0ml=0.084 mol/L *10ml 10^-3 L/ml=0.84 *10^-3 moles

moles of MOPs left after neutralization with strong base=10^-3 moles-0.84 *10^-3 moles=0.16*10^-3 moles

moles of L added=4.69*10^-6 mol/L*1.00 ml* 10^-3 L/ml=4.69*10^-9 moles(base)

L reacts with HA thus,

now,L+HA=LH+A- (net rxn),knet

we know, L+H2O=LH+OH- kb=8.7*10^-7

kb=[LH][OH-]/[L]

HA+H2O=A- +H3O+ ka=6.3*10^-8

ka=[A-] [H3O+]/[HA]

ka*kb=  [A-] [H3O+]/[HA]     [LH][OH-]/[L]=[LH][A-]/[L][HA]*kw

knet =ka*kb/kw=[LH][A-]/[L][HA]

knet =(6.3*10^-8)(8.7*10^-7)/10^-14=54.81 *10^-1=5.48

ICE table

                                     [L]            [HA]                    [LH]           [A-]

Initial conc(moles)       4.69*10^-9    0.16*10^-3             0               0

change                            -X                 -X                        +X             +X

eqm                               4.69*10^-9-X  0.16*10^-3-X x x

knet= [LH][A-]/[L][HA]=X^2/(4.69*10^-9-X) (0.16*10^-3-X)=X^2/(4.69*10^-9) (0.16*10^-3)[x<<small)

5.48=X^2/(4.69*10^-9) (0.16*10^-3)=

4.13 *10^-12=X^2

X=2.03 *10^-6 moles=[LH]=[A-]

fraction of [LH]=[LH]/[L]+[LH]+[A-]+[HA]

=(2.03 *10^-6 moles)/   [(4.69*10^-9)+(2.03 *10^-6) +2.03 *10^-6)+ (0.16*10^-3)]

=2.03 *10^-6/[0.16*10^-3](ignoring smaller values)

=12.68*10^-3=0.0127

[LH]=0.0127

%[LH]=0.0127*100=1.27%