A 0.21g sample of sulfuric acid is dissolved completely in sufficient water to make 0.25 litre of final solution. Calculate the hydrogen ion concentration in mol1-1 in this solution and show successive steps in the calculation: If 0.21g in 0.25l then 0.84g in 1l.You multiplied 0.21 g x 1/0.25L (or 0.21g x 4) = 0.84 g in 1 L here. So you have already corrected for the fact that the 0.21 was not in a liter). Add together molar masses 2x1.01 for hydrogen and 96.06 giving 98.08g molar mass. Number of moles in 0.84g is 0.84 divided by 98.08 giving 0.0085 mol.Then you divided the 0.84/molar mass to convert g/L to mols/L. Would concentration then be 0.0085 mol litre-1 in one litre, divide this by 4 to get 0.25l giving 0.0021mol? I have no idea if I have worked this out correctly so any help would be appreciated! This last step beginning with "Would......" was incorrect and the only thing you didn't do was to multiply the 0.0085 x 2 to account for the 2H^+ that ionize. If you do it through and let the numbers stay in the calculator (not rounding between steps) and round at thr end, you get 0.21/0.25L x 1/98.08 x 2 =
and that is the same as
0.21 x 4 x 1/98.08 x 2 =
which gives 0.01713 mols/L.
I hope this helps

This has helpped me a hell of alot, but i am confused about what the "pH" would be of the sulfuric acid solution

i have ph=2.3_____

= pH 2 (neartest whole number)

i got ph 2 for the answer as well.

how did you get 2.3? I had 1.7x10-2 and ph is the value of the-2.

pH = -log(2*2.14*10^-3)

= 2.36858......

= 2pH

I think both of you are confused a little. Some of the above post was written by a student with my bold face remarks inserted later. With coppying, the bold face comments don't show and it is hard to know what the student wrote initially and what I wrote later. At any rate,
(H^+)=1.713E-2 (note that you must round to the appropriate number of significant figures.)
pH = - log(H^+) = -log(1.713E-2) = -(-1.766)=1.766 which I would round to 1.8.
To the nearest whole number pH = 2.

The (H^+) is NOT 2*2.14*10^-3).
0.21/98.08 = mols H2SO4.
mols H2SO4/0.25L = mols H2SO4/L = M
mols H2SO4/L x 2 = mols H^+.

All together,
0.21/98.08 x 1/0.25L x 2 = 1.713 x 10^-2 M
pH = -log(H^+) = see below.

In (H^+) what is the ^ representing??

Answered above.

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