Calculate the pH of 0.010 M HBr in 0.090 M KBr. actually i solved in two differe

Question

Calculate the pH of 0.010 M HBr in 0.090 M KBr. actually i solved in two different ways because in the first way WHICH I THINK I'M DOING RIGHT BUT HAVE WRONG ANSWER, so i tried the other way WHICH I THINK I'M DOING WRONG AND I DON'T KNOW WHY THATS THE SOLUTION BUT I GET THE RIGHT ANSWER..... (the book provides only the answer: 2.08) The first way i tried: Calculate the ionic strength of all species involved I = (0.5)(SUM c*z^2) where c=concentration z=charge I = (0.5)[(0.010x1^2)+(0.010x(-1)^2)+(0.090x1^2)+(0.090x(-1)^2)] = 0.10 Get the activity coefficient of H+ (i got it from the table) = 0.83 solve pH using pH= -log ([H+]A) pH= -log[(0.10)(0.83)] = 1.08 (which is a different answer from the real answer) The second way i tried: do : pH=-log[(0.010)(0.83)] = 2.08 (which is the right answer) and i want to ask : don't we need to consider also KBr? where did the KBr go????????? why is the solution like this?? is there any other solutions where you can also get 2.08?? please answer....thank you

Explanation / Answer

KBr is a salt. pH is only due to the hydronium ions' concentration. So, the presence of salt really doesn't matter unless it has got H+ ions to furnish.

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