Question

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 50.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

Answer 1

we have:

Molarity of HNO3 = 0.1 M

Volume of HNO3 = 50 mL

Molarity of NH3 = 0.1 M

Volume of NH3 = 100 mL

mol of HNO3 = Molarity of HNO3 * Volume of HNO3

mol of HNO3 = 0.1 M * 50 mL = 5 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.1 M * 100 mL = 10 mmol

We have:

mol of HNO3 = 5 mmol

mol of NH3 = 10 mmol

5 mmol of both will react

excess NH3 remaining = 5 mmol

Volume of Solution = 50 + 100 = 150 mL

[NH3] = 5 mmol/150 mL = 0.0333 M

[NH4+] = 5 mmol/150 mL = 0.0333 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.745+ log {3.333*10^-2/3.333*10^-2}

= 4.74

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.74

= 9.26

Answer: 9.26