Question

85.      

What is the pH of an aqueous solution made by combining 13.02 mL of a 0.1181 M hydrochloric acid with 38.23 mL of a 0.3183 M solution of ammonia?

Answer 1

we have:

Molarity of HCl = 0.1181 M

Volume of HCl = 13.02 mL

Molarity of NH3 = 0.3183 M

Volume of NH3 = 38.23 mL

mol of HCl = Molarity of HCl * Volume of HCl

mol of HCl = 0.1181 M * 13.02 mL = 1.5377 mmol

mol of NH3 = Molarity of NH3 * Volume of NH3

mol of NH3 = 0.3183 M * 38.23 mL = 12.1686 mmol

We have:

mol of HCl = 1.5377 mmol

mol of NH3 = 12.1686 mmol

1.5377 mmol of both will react

excess NH3 remaining = 10.6309 mmol

Volume of Solution = 13.02 + 38.23 = 51.25 mL

[NH3] = 10.6309 mmol/51.25 mL = 0.2074 M

[NH4+] = 1.5377 mmol/51.25 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)

= 4.745

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.745+ log {3*10^-2/0.2074}

= 3.9

we have below equation to be used:

PH = 14 - pOH

= 14 - 3.9

= 10.1

Answer: PH = 10.1