Question

Consider the titration of 29.00 mL of aqueous ammonia with 18.70 mL of 0.1250 M HCl (aq). Kb for NH3 (aq) is 1.8 x 10^-5.

A.) What is the formula of the solute at the equivalence point?

B.) What is the balanced equation for the ionization of the cation of the solute?

C.) What is the molarity of the solute at the equivalence point?

D.) What is the pH at the equivalence point?

Answer 1

You must recognize what is going on in the titration. The equation is
NH3.H2O + HCl ==> NH4Cl + H2O

Obviously NH4Cl is what you have at the equivalence point.
Hydrolyze NH4Cl to determine the pH of th solute.
NH4^+ + H2O ==> H3O^+ + NH3
Set up the ICE chart, and plug all into Ka for the hydrolysis. Ka = (Kw/Kb) =
(NH3)(H3O^+)/(NH4^+)

molarity of solute is moles/L.
Post your work if you get stuck.

Answer 2

I understand the ICE chart, but I'm unsure where you get the values to put in the chart?

Answer 3

This is hydrolysis of 'salt' ions... After titration to equivalence point, only salt remains in solution. Before hydrolysis, only salt ions will be present. Using NH4Cl => NH4+ + Cl- only the NH4+ will under go hydrolysis because theoretically Cl- + HOH => HCL + OH-. However, HCl being a strong acid (electrolyte) prefers to remain 100% ionized. This leaves:
NH4+ + HOH => NH4OH + H+
Ci: [salt] --- 0 0
∆C: -x --- +x +x
Ceq [salt-x]* --- x x
~[salt]
----
*you can drop 'x' under NH4+
----
Apply equilibrium row to Ka expression for Hydrolysis of NH4+.

Ka(NH4+) = (Kw/Kb)
=([NH4OH][H+])/([NH4+)]
=(x)(x)/[NH4+] = x^2/[NH4+]

x = [H+] = ((Kw/Kb)[NH4+])^1/2
pH at equivalence point = -log[H+]