Question

Calculate the pH during the titration of 30.00 mL of 0.1000 M morphine with 0.1000 M HCl(aq) after 9 mL of the acid have been added. K_b of morphine = 1.6 x 10^-6.

Answer 1

Let the morphine be denoted by symbol B

Given:

M(HCl) = 0.1 M

V(HCl) = 9 mL

M(B) = 0.1 M

V(B) = 30 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 9 mL = 0.9 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.1 M * 30 mL = 3 mmol

We have:

mol(HCl) = 0.9 mmol

mol(B) = 3 mmol

0.9 mmol of both will react

excess B remaining = 2.1 mmol

Volume of Solution = 9 + 30 = 39 mL

[B] = 2.1 mmol/39 mL = 0.0538 M

[BH+] = 0.9 mmol/39 mL = 0.0231 M

They form basic buffer

base is B

conjugate acid is BH+

Kb = 1.6*10^-6

pKb = - log (Kb)

= - log(1.6*10^-6)

= 5.796

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 5.796+ log {2.308*10^-2/5.385*10^-2}

= 5.428

use:

PH = 14 - pOH

= 14 - 5.4279

= 8.5721

Answer: 8.57