Acid-Base and pH

[Grimm]

500 mL of 0.20 M solution of Ammonia is mixed with 500 mL of 0.20 M solution of HBr. What is the pH of the resulting solution of NH4Br?

 

Ok, so I know the general math required. I'm just not sure how to proceed here. I'm using mostly the book provided and I'm not seeing an example of how to proceed.

 

.20 M / 500 mL = .40 M

 

 

NH3 + H2O <---> OH- + NH4+

 

HBr + H2O ---> H3O+ + Br-

 

 

I know NH3 is a weak base, and HBr is a strong acid. How do I find the pH of a mixture of two different solutions. Seriously my book says nothing about this. Should I just use the formula for NH4Br? What would the k value of NH4Br be?

 

Any help would be greatly appreciated. Thank you.

[John Cuthber]

Did your course cover this?
https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

[Grimm]

Did your course cover this?

https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equa

Yes, I know that k=products/reactants. This won't help find the concentration of H+ or OH- unless I know the k values first.

If I try using:

 

NH4+ + H2O <---> H3O+ + NH3

 

I get a pH of 4.82

 

The book says the answer is 5.13

 

Wish I knew exactly what I was doing wrong here.

If 500.0 mL of a 0.20 M solution of ammonia is mixed with 500.0 mL of a 0.20 M solution of HBr, what is the pH of the resulting solution of NH4Br ?

[Grimm]

Damn, I thought someone here could help me for sure. Thanks for your time.

[studiot]

 

 

20 M / 500 mL = .40 M

 

So where did this come from?

[Grimm]

 

So where did this come from?

If 500.0 mL of a 0.20 M solution of ammonia is mixed with 500.0 mL of a 0.20 M solution of HBr, what is the pH of the resulting solution of NH4Br ?

 

The above is the question I'm trying to answer. I converted the .20 M because M stands for mol/L and the volume for the question is given in mL. .20 mol /.5 L = .4 M

Edited October 1, 2016 by Grimm

[Fuzzwood]

Ok, so I know the general math required.

Apparently you don't. If you did then you would have figured that M stands for moles/liter and that taking half a liter of solution does NOT double the amount of whatever is dissolved.

[Grimm]

Apparently you don't. If you did then you would have figured that M stands for moles/liter and that taking half a liter of solution does NOT double the amount of whatever is dissolved.

 

Ok I don't know shit then. Will you help me now?

It was my understanding that it doesn't double the actual amount in the container but the concentration would be greater since it's the same amount of material in a smaller container. Please enlighten me if this is not the correct way to approach this.

[studiot]

So at the start of the reaction, what is the volume of the mixture ?

 

An how many moles of hydrogen bromide does it contain?

 

An how many moles of ammonia?

 

So what are the concentrations of each?

[Grimm]

At the start of the reaction the volume of the mixture would have to be 1 L.

 

As for the number of moles of each, n=CV

 

HBr n= (.20 M)(1 L)

HBr n= .20 moles

 

NH3 n= (.20 M)(1 L)

NH3 n= .20 moles

 

So the concentrations of each would have to be [HBr] = .20 mol/L and [NH3] = .20 mol/L

[Grimm]

I would like to announce that I have solved the problem. The only thing I did wrong was the beginning concentrations!

I feel so stupid. I think I was remembering earlier problems that always tried throwing in different amounts of molarities and container sizes to make sure you were paying attention.

M1V1 = M2V2

(.20 mol/L)(.5 L) = M2(1.0 L)

.1 mol = M2(1.0 L)

.1 mol/L = M2

If I use this in my ICE table, I get the right answer. pH=5.13

ka = [H3O+][NH3] / [NH4+]

5.6 x 10^-10 = [x^2] / [.10 - x] With such a small k value I can use the 5% rule.

5.6 x 10^-10 = [x^2] / [.10]

5.6 x 10^-11 = [x^2]

7.48 x 10^-6 = x = [H3O+]

-log(7.48 x 10^-6) = 5.13 = pH

Thanks to everyone that helped!

Edited October 2, 2016 by Grimm

[studiot]

Whilst I'm pleased that my hint helped, I glad that you worked it out for yourself.

That's by far the best way and hopefully you will understand it better now so that it will 'stick'.

+1

[Grimm]

Thank you again studiot! I definitely have to watch my starting concentrations more closely in the future.

 

Laziness and hubris, bad combo.