[daylight]

I think I may have the right idea on how to set this up, but I am not sure I have done it correctly, please help...

 What is the concentration of benzoic acid that is necessary to produce a solution that has a pH of 3.5?(Ka=6.5x10-5)
C₆H₅COOH = H₂O  H₃O+ + C₆H₅COO-

[H30+}=10^-3.5 = 3.16X10^-4
X= MOL BENZOIC ACID/L

Ka=6.5x10-5=[H₃O+][C₆H₅COO-]/ [C₆H₅COOH]
X=(3.16X10^-4)²/3.16X10^-4
I am not sure if I should set this = to the value of Ka now? and solve for x or just solve for x and Ka doesn't fit in? I would think I need to use Ka? also, does the bottom (3.16 x 10-4)concentration cancel out and I am left with (3.16x10-4)/x=Ka?

hope this makes some sense....

[Borek]

I think I may have the right idea on how to set this up, but I am not sure I have done it correctly, please help...

 What is the concentration of benzoic acid that is necessary to produce a solution that has a pH of 3.5?(Ka=6.5x10-5)
C₆H₅COOH = H₂O  H₃O+ + C₆H₅COO-

[H30+}=10^-3.5 = 3.16X10^-4
X= MOL BENZOIC ACID/L

Ka=6.5x10-5=[H₃O+][C₆H₅COO-]/ [C₆H₅COOH]

So far so good, but

X=(3.16X10^-4)²/3.16X10^-4

This is wrong. [H₃O⁺] is the same as [C₆H₅COO^-] but different from [C₆H₅COOH]. And don't forget that what you are looking for is a sum of [C₆H₅COO^-]+[C₆H₅COOH].

[daylight]

Oh, I think I left out the x- on the bottom? It should read /x-3.16x10^-4 correct?

[Borek]

Depends on what do you mean by x.

[daylight]

X would be the concentration of benzoic acid that I am looking for. I've tried working it out that way
Ka=6.5x10-5=[H3O+][C6H5COO-]/ [C6H5COOH]
6.5x10-5=(3.16X10-4)2/x-(3.16X10-4)

6.5x10-5=(3.16X10-4)/x
x=(3.16X10-4)/6.5x10-5 = 4.86=[CgH5Cooh](since x-=(3.16X10-4) is basically the same as x.
??

[Borek]

6.5x10-5=(3.16X10-4)2/x-(3.16X10-4)

6.5x10-5=(3.16X10-4)/x

You have lost me here, there was a ² in the upper equation, where is it now?

As I have already signalled earlier, your x is not amount of benzoic acid you should use.

[daylight]

6.5x10-5=(3.16X10-4)2/x-(3.16X10-4)

6.5x10-5=(3.16X10-4)/x

You have lost me here, there was a ² in the upper equation, where is it now?

As I have already signalled earlier, your x is not amount of benzoic acid you should use.

I took =(3.16X10-4)2/x-(3.16X10-4) and divided it out, (which left me with 3.16x10-4/x so I multipled 6.5x10-5 by x and then divided by 3.16x10-4.
So are you saying the bottom half should not be x-3.16x10-4? my algebra skills are definetly not too hot, so that could be part of the problem (I haven't had algebra in 20 years, so bear with me, it's my weakest subject).

"And don't forget that what you are looking for is a sum of [C6H5COO-]+[C6H5COOH]." so is this where I am missing it?  is the concentration just the left side added together and not to be factored out of the Ka=concentrations? I'm sorry I'm not getting it quicker, but chemistry is hard for me and I really need things pointed out sometimes even when the look obvious.

[Borek]

I took =(3.16X10-4)2/x-(3.16X10-4) and divided it out, (which left me with 3.16x10-4/x so I multipled 6.5x10-5 by x and then divided by 3.16x10-4.

Something like the attached picture? That's algebraically incorrect, you can't cancel out things that are part of the sum in the denominator with something that is in the nominator.

[daylight]

If I divide A2 by A I get A, but it's negative, so I thought maybe it was wrong, but then what else do I do? do I multiply (x-3.16x10-4) by 6.5x10-4?

[Borek]

6.5x10^-5=x²/(x-3.16x10^-4)

You have to multiply both sides by (x-3.16x10^-4) first. Note, that you will get quadratic equation, that can't be simply and directly solved form x.

http://en.wikipedia.org/wiki/Quadratic_equation