Question

What is the pH of a solution that is produced by mixing 0.6 liters of a 0.02M solution ammonia and 0.2 liters of 0.015M HBr.

Please show all work as I would like to learn how to do the problem and not just have the answer. Thank you.

Answer 1

Given:

M(HBr) = 0.015 M

V(HBr) = 0.2 L

M(NH3) = 0.02 M

V(NH3) = 0.6 L

mol(HBr) = M(HBr) * V(HBr)

mol(HBr) = 0.015 M * 0.2 L = 0.003 mol

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

mol(NH3) = 0.02 M * 0.6 L = 0.012 mol

We have:

mol(HBr) = 0.003 mol

mol(NH3) = 0.012 mol

0.003 mol of both will react

excess NH3 remaining = 0.009 mol

Volume of Solution = 0.2 + 0.6 = 0.8 L

[NH3] = 0.009 mol/0.8 L = 0.0113 M

[NH4+] = 0.003 mol/0.8 L = 0.0037 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 {3.75*10^-3/1.125*10^-2}

= 4.268

use:

PH = 14 - pOH

= 14 - 4.2676

= 9.7324

Answer: 9.73