Question

Calculate the pH of a solution that results upon mixing 40ml of 0.09M NH3 with 8ml of 0.18M HNO3

Answer 1

we have:

Molarity of HNO3 = 0.18 M

Volume of HNO3 = 8 mL

Molarity of NH3 = 0.09 M

Volume of NH3 = 40 mL

Kb for NH3 = 1.8*10^-5

mol of HNO3 = Molarity of HNO3 * Volume of HNO3

mol of HNO3 = 0.18 M * 8 mL = 1.44 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.09 M * 40 mL = 3.6 mmol

We have:

mol of HNO3 = 1.44 mmol

mol of NH3 = 3.6 mmol

1.44 mmol of both will react

excess NH3 remaining = 2.16 mmol

Volume of Solution = 8 + 40 = 48 mL

[NH3] = 2.16 mmol/48 mL = 0.045 M

[NH4+] = 1.44 mmol/48 mL = 0.03 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)

= 4.7447

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.7447+ log {0.03/0.045}

= 4.5686

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.5686

= 9.43

Answer: 9.43