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. does the # sound good? Im not sure I used the write method to solve it..

Answer 1

I do agree with the first two processes, but the last one, i do not based on one reason, you do not get the pH by getting the log of the number of moles.

To avoid confusion, let me explain it in detail, including the reason for the processes involved.

You are given a strong acid and a weak base.

First thing to do is to calculate the reaction as a limiting reagent problem. To do this, you have to get the number of moles of both the acid and base. This is done by multiplying the concentration and the volume.

nb= (0.250 M NH3)(0.025L)
nb=0.00625 mol NH3

na= (0.250 M HCl)(0.055L)
na = 0.01375 mol HCl

if na = 0 we have a pure weak base
if na if na = nb we have a hydrolysis
In this case na>nb, this means that we have an excess of strong acid.

This means that the reaction of HCl produces nb moles of conjugate acid, which is NH4Cl, this therefore leaves 0 mols of unreacted weak base, and leaves na - nb mols of unreacted strong acid.

Thus, the [H+] is will be equal to the concentration of the unreacted strong acid.

[H+] = (na - nb)/(va+vb)
[H+] = {(0.01375 - 0.00625)/(0.025L+0.055L)
[H+] = 0.09375

pH = - log [H+]
pH = - log (0.09375)
pH = 1.03

Answer 2

The 'sturdy' in 'sturdy acid' in simple terms ability it quite often ionizes. Even a vulnerable acid, one that quite often would not ionize, will react with ammonia. you're able to have considerable extra ammonia, so it quite is going to be alkaline

Answer 3

NH3 (aq) + HCl (aq) ------> NH4Cl (aq)

1) C1V1 = moles of NH3 = (0.025 L)(0.250 mol/L)=0.00625 mol NH3

C2V2= moles of HCl = (0.055 L)(0.250 mol/L) = 0.01375 mol HCl

2) pH of HCl solution is:

pH = -log [H+] = - log (0.01375 mol/L) = 1.86....before adding NH3

3) According to reaction above, adding the ammonia we just neutralize 0.00625 moles of HCl, so,

Remaining HCl moles = 0.01375 moles - 0.00625 moles = 0.0075 moles

4) We calculate again the pH:

final pH = - log (0.0075) = 2.12

Hope other people can confirm results.

Good luck!

P.S. ds_8615 is right, I just took the amount of remaining moles of HCl without regarding the added volume of NH3 to solution of HCl.
I will correct this as follows:

Remaining HCl moles = 0.0075 moles
Divide 0.0075 moles by (0.025 mL + 0.055 mL) = 0.09375 moles

Now pH = - log (0.09375 mol/L) = 1.028

pH = 1.028

Sorry for this mistake. Thanks anyway for letting me in!