A 100.0 mL sample of 0.05 M NH3 is titrated with 0.10 M HCl. Determine the pH of the solution after the addition of 50.0 mL HCl.
Here is my work:
NH3 + H3O+ --> NH4+ + H2O
Before Addition:
NH3= 0.005 mol
H3O= 0 mol
NH4+ = 0 mol
Addition:
NH3 = 0.005 mol
H3O+ = 0.005 mol
NH4+ = 0.005 mol
After Addition:
NH3 = 0 mol
H3O+ = 0 mol
NH4+ = 0.005
ICE Table
NH4+ + H2O --> NH3 + H3O+
INITIAL:
NH3= 0.0333 M (0.005 mol/(0.100 L + 0.0500 L)
NH3 = 0 M
H3O+ = 0 M
CHANGE (respectively):
-x
+x
+x
EQUILIBRIUM (respectively):
0.0333 - x
x
x
Ka expression:
(x^2)/(0.0333-x) = (1 x 10^-14)/(1.8 x 10^-5)
assume x is small approximation...
x= 4.30 x 10^-6
pH= -log (4.30 x 10^-6) = 5.37
FINAL ANSWER: pH = 5.37
Please check.
1 answer
I didn't go through your work step by step but I worked the problem from scratch. Using 1.8E-5 for Kb, I obtained the same answer you did. Good work.
1 answer