Find pH for the solution resulting in adding ammonia and HCl?

Question

Calculate the pH of the solution resulting from the addition of 25.0 mL of 0.250M ammonia and 55.0 mL of 0.250 M HCl solution.

I got 5.18 for the pH but im not so certain I did it right, can anyone else solve this and tell me what you got.

Answer 1

Find the moles n1 of NH3 and the moles n2 of HCl

n1 = C1*V1 = 0.25*0.025 = 6.25*10^(-3) mol NH3
n2 = C2*V2 = 0.25*0.055 = 0.01375 mol HCl

Find the final volume: Vf = V1 + V2 = 0.025 + 0.055 = 0.08 L

Now, HCl is in excess, NH3 reacts fully:

NH3 + HCl --> NH4Cl

So the final solution contains: n = 0.01375 - 6.25*10^(-3) = 7.5*10^(-3) mol HCl

Find the molarity of HCl in the final solution:

C = n/Vf = 7.5*10^(-3)/0.08 = 0.09375 M

HCl is a strong acid, it reacts fully with water:

HCl + H2O --> H3O+ + Cl-

So [H3O+] = 0.09375M and pH = -log[H3O+], pH = -log(0.09375),

pH = 1.03 or pH = 1 approx.

Answer 2

The 'solid' in 'solid acid' merely skill it as a rule ionizes. Even a susceptible acid, one that as a rule would not ionize, will react with ammonia. you have substantial extra ammonia, so it incredibly is going to likely be alkaline