[slayer]

How many moles of C₂H₅CO₂H (aq) are needed to make 500.0 mL of a solution with a pH of 3.33?

this is the dominant equilibrium to be used in the reaction:
C₂H₅CO₂H (aq) + H₂O (l) <--> H₃O⁺ (aq) + [C₂H₅CO₂^- (aq) where Ka = 1.3 x 10^-5

The first thing that I did was determine the concentration for a pH solution of 3.33

pH = 3.33
pOH = 14 - 3.33
pOH = 10.67
-logx = 10.67
x = 10^-10.67
x = 2.13796 x 10^-11M

then, I constructed an ice table:
C₂H₅CO₂H (aq) + H₂O (l) <--> H₃O⁺ (aq) + C₂H₅CO₂^- (aq)
I)    2.13796 x 10^-11 + water <---> 0 + 0
C)   -x + water <---> +x + +x
E)   2.13796 x 10^-11 - x + water <---> x + x

now i evaluate the ice table results:
Keq = [H₃O⁺] [C₂H₅CO₂^-] / [C₂H₅CO₂H]
1.3 x 10^-5 = [x²] / [2.13796 x 10^-11 - x] (then we throw away the bottom x)
1.3 x 10^-5 * 2.13796 x 10^-11 = x²
x² = 2.779 x 10^-16
x = 1.67 x 10^-8M

then we convert the volume from mL to L to derive the moles:
500mL * [1L / 1000mL] = .500L

now we derive the moles:
1.67 x 10^-8M = 1.67 x 10^-8 mol / L * .500L = 8.34 x 10^-9

is this correct?
(thanks alot for your responses.. i never realized how hard it is to write these things! )

[Borek]

The first thing that I did was determine the concentration for a pH solution of 3.33

pH = 3.33
pOH = 14 - 3.33
pOH = 10.67
-logx = 10.67
x = 10^-10.67
x = 2.13796 x 10^-11M

And this is concentration of what?

So sad you have spent so much time formatting whole post, yet you have made an error at the very beginning...

Try to start not with ICE table, but with the most simplified equation (as shown here: equation 8.13). It is valid only if the dissociation fraction is below 5% (see table down the page), so you will have to check if the condition is meet - but if so, calculations are pretty fast. If not, go for ICE.

[slayer]

borek,

do I use the pH = 3.33
-logx = 3.33
x = 10^-3.33??

because we are producing H₃O⁺?

and this would be the concentration of the acid.

is this correct?

[Borek]

borek,

do I use the pH = 3.33
-logx = 3.33
x = 10^-3.33??

because we are producing H₃O⁺?

Yes.

and this would be the concentration of the acid.

No. You have to use this value (H⁺ concentration) to find out what IS the weak acid concentration.

[slayer]

then, I constructed an ice table:
C₂H₅CO₂H (aq) + H₂O (l) <--> H₃O⁺ (aq) + C₂H₅CO₂^- (aq)
I)    2.13796 x 10^-11 + water <---> 0 + 0
C)   -x + water <---> +x + +x
E)   2.13796 x 10^-11 - x + water <---> x + x

So my ice table is actually this:
C₂H₅CO₂H (aq) + H₂O (l) <--> H₃O⁺ (aq) + C₂H₅CO₂^- (aq)
I) 0 + water <--> 4.68 x 10^-4 + 0
C) +x + water <--> -x + +x
E) x <--> (4.68 x 10^-4 - x) + x
therefore; Keq = ([Hydronium][C₂H₅CO₂^-]) / [C₂H₅CO₂H]...

therefore
Keq = (4.68 x 10^-4 -x * x) / x ?

[Borek]

So my ice table is actually this:
C₂H₅CO₂H (aq) + H₂O (l) <--> H₃O⁺ (aq) + C₂H₅CO₂^- (aq)
I) 0 + water <--> 4.68 x 10^-4 + 0

Why 0 for initial C₂H₅COOH?

C) +x + water <--> -x + +x

why +x for change of C₂H₅COOH? -x for H⁺?

Slow down and try to think - what is what of what

[slayer]

4.68 x 10^-4M must be the concentration of H₃O⁺ necessary to have that pH.

0 for initial C₂H₅COOH because I am adding it to get to the 3.33 pH, right?

and the +x for C₂H₅COOH because I need to find out how much of it i need to add?

[Borek]

4.68 x 10^-4M must be the concentration of H₃O⁺ necessary to have that pH.

OK

0 for initial C₂H₅COOH because I am adding it to get to the 3.33 pH, right?

No. You are calculating pH of solution that contains concentration C.

and the +x for C₂H₅COOH because I need to find out how much of it i need to add?

No. Normally you will do ICE to calculate pH. You have to reverse the approach. Do standard ICE for concentration C, insert known pH, then solve for C.

[slayer]

borek. im lost.

i dont think im supposed to be adding and subtracting x's because I already determined what my X is from the pH calculation.

when you say concentration C, are you implying I need to use the C1V1 = C2V2 eq?

No. Normally you will do ICE to calculate pH. You have to reverse the approach. Do standard ICE for concentration C, insert known pH, then solve for C.

i dont know how to approach this

[Borek]

HA <-> H⁺ + A^-

standard ICE, with C standing for analytical concentration of acid:

I      C        0        0
C     -x       +x      +x
E     C-x     x        x

Ka = x²/(C-x)

Usually you solve for x - this time insert x (10^-pH) and solve for C. You are asked about C, and you are given pH, aren't you?

[Donaldson Tan]

standard ICE, with C standing for analytical concentration of acid:

LOL. You are not using your usually recommended HH method.

[Borek]

LOL. You are not using your usually recommended HH method.

I am not nailed to any method While I prefer some over other, I can use other ones as well, when I feel like it makes sense. slayer was lost in his own approach so it was good for him to show where he went astray.