Grimm Posted September 30, 2016 Share Posted September 30, 2016 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. Link to comment Share on other sites More sharing options...
John Cuthber Posted September 30, 2016 Share Posted September 30, 2016 Did your course cover this?https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation Link to comment Share on other sites More sharing options...
Grimm Posted October 1, 2016 Author Share Posted October 1, 2016 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 ? Link to comment Share on other sites More sharing options...
Grimm Posted October 1, 2016 Author Share Posted October 1, 2016 Damn, I thought someone here could help me for sure. Thanks for your time. Link to comment Share on other sites More sharing options...
studiot Posted October 1, 2016 Share Posted October 1, 2016 20 M / 500 mL = .40 M So where did this come from? 1 Link to comment Share on other sites More sharing options...
Grimm Posted October 1, 2016 Author Share Posted October 1, 2016 (edited) 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 Link to comment Share on other sites More sharing options...
Fuzzwood Posted October 1, 2016 Share Posted October 1, 2016 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. 2 Link to comment Share on other sites More sharing options...
Grimm Posted October 1, 2016 Author Share Posted October 1, 2016 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. Link to comment Share on other sites More sharing options...
studiot Posted October 1, 2016 Share Posted October 1, 2016 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? 1 Link to comment Share on other sites More sharing options...
Grimm Posted October 1, 2016 Author Share Posted October 1, 2016 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 Link to comment Share on other sites More sharing options...
Grimm Posted October 2, 2016 Author Share Posted October 2, 2016 (edited) 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 1 Link to comment Share on other sites More sharing options...
studiot Posted October 2, 2016 Share Posted October 2, 2016 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 Link to comment Share on other sites More sharing options...
Grimm Posted October 2, 2016 Author Share Posted October 2, 2016 Thank you again studiot! I definitely have to watch my starting concentrations more closely in the future. Laziness and hubris, bad combo. Link to comment Share on other sites More sharing options...
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