Question

When 50.0mL of 0.100M NH3 (Kb=1.8x10^-5) solution is mixed with 20.0mL of of 0.250M HCl solution, what is the pH after reaction?

I solved the ka=5.56x10^-10 and I solved the moles of each nHCl and nNH3 and both have 0.005 moles. What do I do after this?

Thanks.

Answer 1

•Moles of NH3 = (0.0500L)(0.100mol/L) = 0.00500 mol NH3.
•0.0500 L + 0.0200 L = 0.0700 L total volume
•Moles of NH3/Liter = 0.00500 mol NH3/0.0700 L) = 0.07143 mol/L = 0.07143 M NH3 (before reacting with HCl)
•(0.250mol/L)(0.0200L)/(0.0700L) = 0.07143 M HCl
The reaction is:
NH3(aq) + HCl(aq) --> NH4+(aq) + Cl-(aq)
The reaction produces NH4+ , a week acid with the same concentration as the inititial concentrations of NH3 or HCl, 0.07143 M. The equilibrium for the newly formed NH4+ is:
NH4+(aq) <=> NH3(aq) + H+(aq)
Ka = [NH3][H+] / [NH4+]
Let [H+] = [NH3] = x
<b>Ka = (x)(x) / [NH4+]</b>
•Look up the Ka for NH4+
•[NH4+] = 0.07143M
•Substitute, solve for x
•pH = -log(x)