Question

An analytical chemist is titrating 228.9 mL of a 0.6800 M solution of isopropylamine ((CH3)2CHNH2) with a 0.5500 M solution of HNO3. The pKb of isopropylamine is 3.33. Calculate the pH of the base solution after the chemist has added 249.4mL of the HNO3 solution to it.  

Answer 1

use:
pKb = -log Kb
3.33= -log Kb
Kb = 4.677*10^-4
Given:
M(HNO3) = 0.55 M
V(HNO3) = 249.4 mL
M((CH3)2CHNH2) = 0.68 M
V((CH3)2CHNH2) = 228.9 mL


mol(HNO3) = M(HNO3) * V(HNO3)
mol(HNO3) = 0.55 M * 249.4 mL = 137.17 mmol

mol((CH3)2CHNH2) = M((CH3)2CHNH2) * V((CH3)2CHNH2)
mol((CH3)2CHNH2) = 0.68 M * 228.9 mL = 155.652 mmol



We have:
mol(HNO3) = 137.17 mmol
mol((CH3)2CHNH2) = 155.652 mmol

137.17 mmol of both will react
excess (CH3)2CHNH2 remaining = 18.482 mmol
Volume of Solution = 249.4 + 228.9 = 478.3 mL
[(CH3)2CHNH2] = 18.482 mmol/478.3 mL = 0.0386 M
[(CH3)2CHNH3+] = 137.17 mmol/478.3 mL = 0.2868 M

They form basic buffer
base is (CH3)2CHNH2
conjugate acid is (CH3)2CHNH3+

Kb = 4.677*10^-4

pKb = - log (Kb)
= - log(4.677*10^-4)
= 3.33

use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.33+ log {0.2868/3.864*10^-2}
= 4.201

use:
PH = 14 - pOH
= 14 - 4.2005
= 9.7995

Answer: 9.80