Question

1. (1 point) Calculate the pH of a solution obtained by adding 12.0 mL of 0.25 M HCl to 6.0 mL of 1.0 M NH₃.

Answer 1

Given:
M(HCl) = 0.25 M
V(HCl) = 12 mL
M(NH3) = 1 M
V(NH3) = 6 mL


mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.25 M * 12 mL = 3 mmol

mol(NH3) = M(NH3) * V(NH3)
mol(NH3) = 1 M * 6 mL = 6 mmol



We have:
mol(HCl) = 3 mmol
mol(NH3) = 6 mmol

3 mmol of both will react
excess NH3 remaining = 3 mmol
Volume of Solution = 12 + 6 = 18 mL
[NH3] = 3 mmol/18 mL = 0.1667 M
[NH4+] = 3 mmol/18 mL = 0.1667 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 {0.1667/0.1667}
= 4.745

use:
PH = 14 - pOH
= 14 - 4.7447
= 9.2553

Answer: 9.26