posted by bert on .
Consider the titration of 50 mL of 0.15 M NH3 with 0.1 M HCl. What is the pH at the equivalence point of the titration? (Kb= 1.8X 10-5 Ka= 5.6X10-10)
The salt present at the equivalence point is NH4Cl and there are 0.05 L x 0.15 M. That hydrolyzes at the equivalence point and that determines the pH.
NH4^+ + HOH ==> NH3^+ + H3O^+
Ka = Kw/Kb = (NH3)(H3O^+)/(NH4^+)
Plub in (NH4^+) from mols NH4Cl/liters (don't forget in liters to add the volume of NH3 initially (0.050 L) to the volume of HCl used in the titration.) and solve for H3O^+.
pH = - log (H^+). Post your work if you get stuck.
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