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 0.074 M NaOH . 0.074 M NaOH. Next, you add 1.00 mL of 3.51 × 10 − 5 M 3.51×10−5 M lidocaine to this mixture. Denoting lidocaine as L, calculate the fraction of lidocaine present in the form LH +

Answer 1

Answer:

The fraction of lidocaine is 0.999

Explanation:

The number of moles of MOPS:

$n_{MOPS} =0.01*0.1=1x10^{-3} moles$

The number of moles of NaOH:

$n_{NaOH} =0.01*0.074=7.4x10^{-4} moles$

In 20 mL of solution, the molarity is:

$M_{NaOH}=\frac{7.4x10^{-4} }{0.02} =0.037mol/L$

The acid form is:

1x10⁻³ - 7.4x10⁻⁴ = 2.6x10⁻⁴moles

$M_{MOPS}=\frac{2.6x10^{-4} }{0.02} =0.013mol/L$

$pKa=-logKa=-log(3.51x10^{-5} )=4.45$

$pH=pKa+log\frac{[NaOH]}{[MOPS]} =4.45+log\frac{0.037}{0.013} =4.9$

About lidocaine, the pKa is:

Ka = 1x10⁻¹⁴/8.7x10⁻⁷=1.15x10⁻⁸

$pKa=-logKa=-log(1.15x10^{-8} )=7.94$

$pH=pKa+log\frac{[base]}{[acid]} \\4.9=7.94+log\frac{[base]}{[acid]}\\log\frac{[base]}{[acid]}=-3.04\\base/acid=9.12x10^{-4}$

The fraction of lidocaine is:

$f=\frac{1}{1.009} =0.999$