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.

•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
Ka = (x)(x) / [NH4+]
•Look up the Ka for NH4+
•[NH4+] = 0.07143M
•Substitute, solve for x
•pH = -log(x)

To find the pH after the reaction, you will need to calculate the concentration of the resulting solution and then use the appropriate equation to determine the pH. Here's how you can proceed:

1. Calculate the initial moles of NH3 and HCl:
Moles of NH3 = 0.100 M * 0.050 L = 0.005 mol
Moles of HCl = 0.250 M * 0.020 L = 0.005 mol

2. Since both NH3 and HCl react in a 1:1 ratio, their moles will completely react with each other. This means that the moles of NH3 remaining will be zero after the reaction.

3. The moles of excess HCl (0.005 mol) will react with water to form H3O+ ions: HCl + H2O → H3O+ + Cl-

4. Calculate the volume of the resulting solution:
Total volume = initial volume of NH3 + initial volume of HCl = 50.0 mL + 20.0 mL = 70.0 mL = 0.070 L

5. Calculate the concentration of H3O+ ions formed from the reaction of HCl with water:
Concentration of H3O+ = Moles of HCl / Total volume = 0.005 mol / 0.070 L = 0.071 M

6. Use the concentration of H3O+ to find the pH:
pH = -log[H3O+]
pH = -log(0.071)
pH ≈ 1.15

Therefore, the pH of the solution after the reaction is approximately 1.15.