Question

A solution was prepared from 10.0 mL of 0.100 M cacodylic acid (HA – Ka = 6.4 x 10-7 ) and 10.0 mL of 0.0800 M NaOH. To this mixture was added 1.00 mL of 1.27 x 10-6 M morphine (B – Kb = 1.6 x 10-6 ). What is the fraction of morphine present in the BH+ form?

Please solve in details

Thanks in advance

Answer 1

The buffer prepared from cacodylic acid:

0.010L x 0.100 mol/L = 1 mmol HA

0.010L x 0.0800 mol/L = 0.8 mmol NaOH

After neutralization:

[HA] = 0.2 mmol

[A-] = 0.8 mmol , both in 20 mL sol

pKa = - log6.4 x 10-7 = 6.20

pH = pKa + log ([A-]/[HA]) = 6.2 + log (0.8/0.2) = 6.8

In this buffer solution, the added quantity of morphine is very small (1.27x10^-9 mol) (the morphine equilibration is not consuming the buffer in a significant proportion)

.

pKb of morphine = 5.8, and its pKa is 14 –pKb = 8.2

The ratio [B]/[BH+] obey the same formula as for the buffer

pH = pKa + log([B]/[BH+])

6.8 = 8.2 + log([B]/[BH+])

log([B]/[BH+]) = - 1.4

[B]/[BH+] = 10^-1.4 = 0.04 or 4:100

The fraction is   100 x 100/(100+4) = 96 %