Question

If you titrate 10.0 mL of 0.25 M morpholine (C4H8ONH, pKb = 5.51) with 0.10 M HNO3, what would be the pH after addition of 20.0 mL of HNO3 solution?

Answer 1

we have below equation to be used:

pKb = -log Kb

5.51= -log Kb

log Kb = -5.51

K = 10^(-5.51)

Kb = 3.09*10^-6

we have:

Molarity of HNO3 = 0.1 M

Volume of HNO3 = 20 mL

Molarity of C4H8ONH = 0.25 M

Volume of C4H8ONH = 10 mL

mol of HNO3 = Molarity of HNO3 * Volume of HNO3

mol of HNO3 = 0.1 M * 20 mL = 2 mmol

mol of C4H8ONH = Molarity of C4H8ONH * Volume of C4H8ONH

mol of C4H8ONH = 0.25 M * 10 mL = 2.5 mmol

We have:

mol of HNO3 = 2 mmol

mol of C4H8ONH = 2.5 mmol

2 mmol of both will react

excess C4H8ONH remaining = 0.5 mmol

Volume of Solution = 20 + 10 = 30 mL

[C4H8ONH] = 0.5 mmol/30 mL = 0.0167 M

[C4H8ONH+] = 2 mmol/30 mL = 0.0667 M

They form basic buffer

base is C4H8ONH

conjugate acid is C4H8ONH+

Kb = 3.09*10^-6

pKb = - log (Kb)

= - log(3.09*10^-6)

= 5.51

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 5.51+ log {6.667*10^-2/1.667*10^-2}

= 6.11

we have below equation to be used:

PH = 14 - pOH

= 14 - 6.11

= 7.89

Answer: 7.89