Question

What is the pH of a solution prepared by mixing 50.00mL of 0.12M NH3 and 3.50mL of 1.0M HCl

Answer 1

we have:

Molarity of HCl = 1 M

Volume of HCl = 3.5 mL

Molarity of NH3 = 0.12 M

Volume of NH3 = 50 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 1 M * 3.5 mL = 3.5 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.12 M * 50 mL = 6 mmol

We have:

mol of HCl = 3.5 mmol

mol of NH3 = 6 mmol

3.5 mmol of both will react

excess NH3 remaining = 2.5 mmol

Volume of Solution = 3.5 + 50 = 53.5 mL

[NH3] = 2.5 mmol/53.5 mL = 0.0467 M

[NH4+] = 3.5 mmol/53.5 mL = 0.0654 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 {0.0654/0.0467}

= 4.89

we have below equation to be used:

PH = 14 - pOH

= 14 - 4.89

= 9.11

Answer: 9.11