Question

A 28.9 mL sample of 0.255 M ammonia, NH3, is titrated with 0.309 M hydrochloric acid. After adding 8.68 mL of hydrochloric acid, the pH is

Please show work

Answer 1

Given:

M(HCl) = 0.309 M

V(HCl) = 8.68 mL

M(NH3) = 0.255 M

V(NH3) = 28.9 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.309 M * 8.68 mL = 2.6821 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.255 M * 28.9 mL = 7.3695 mmol

We have:

mol(HCl) = 2.6821 mmol

mol(NH3) = 7.3695 mmol

2.6821 mmol of both will react

excess NH3 remaining = 4.6874 mmol

Volume of Solution = 8.68 + 28.9 = 37.58 mL

[NH3] = 4.6874 mmol/37.58 mL = 0.1247 M

[NH4+] = 2.6821 mmol/37.58 mL = 0.0714 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

use:

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

= 4.745+ log {7.137*10^-2/0.1247}

= 4.502

use:

PH = 14 - pOH

= 14 - 4.5023

= 9.4977

Answer: 9.50