Chemistry Question Not making sense need help asap (1 Viewer)

ebbygoo

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Hey guys.

HSC 2003, question is:

A sulfuric acid solution has a concentration of 5 x 10^4 mol/L

What is the pH assuming completely ionised

The options are
a) 3.0
b) 3.3
c) 3.6
d) 4.0

Using the formula, pH = -log[H+], I get b, but the answer is a?

Please explain? The only thing I can think of is sig figs, but even if so, then shouldn't it be 3 rather than 3.0?

So ye, trial tomoz, help appreciated.

kthxbai
 

photastic

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Hey guys.

HSC 2003, question is:

A sulfuric acid solution has a concentration of 5 x 10^4 mol/L

What is the pH assuming completely ionised

The options are
a) 3.0
b) 3.3
c) 3.6
d) 4.0

Using the formula, pH = -log[H+], I get b, but the answer is a?

Please explain? The only thing I can think of is sig figs, but even if so, then shouldn't it be 3 rather than 3.0?

So ye, trial tomoz, help appreciated.

kthxbai
Yep answer is a because -log(1x10^-3)=3.0
And about the sig figs, who cares, it's multiple choice :p

Full Working Out:
H2SO4 -> 2H(+) +SO4(2-)
Since sulfuric acid is diprotic and will ionise to produce twice the concentration of H+

therefore [H+] = 2 X (5 X 10^-4) = 1 X 10^-3 Mol/L
therefore pH = -log(1 X 10^-3) = 3.0

Good luck for tmmr, doing it as well tmmr but i'm dropping it anyway so yolo.
 
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the_lebtalian

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its because sulfuric acid ionises twice in water:

H2CO3 -> HCO3^- + H30^+
HCO3^- <> CO3^- + H30^+

hence when you put it in the formula you times the concentration of H+ by 2

-log(5x10^-4 x 2) = 3 :smile:

EDIT: just realised the guy above me edited his post while i was writing mine hahaha
 
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photastic

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its because sulfuric acid ionises twice in water:

H2CO3 <> HCO3^- + H30^+
HCO3^- <> CO3^- + H30^+

hence when you put it in the formula you times the concentration of H+ by 2

-log(5x10^-4 x 2) = 3 :smile:
The first equilibrium arrow is wrong because sulfuric acid completely ionises, obviously since it's a strong acid.
 
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Sulfuric acid is diprotic and releases two hydrogen ions, so you multiply the concentration by 2 and then do logs.

H2SO4 + H2O -----> HSO4- + H3O+

HSO4- + H2O <------> SO4(2-) + H3O+
 

ebbygoo

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Whoa idk how I missed this hahhaa.

Ty guys.

Gl to y'all too.
 

faker

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I was under the impression that all acids disassociations are under equilibrium ? Markers I believe shouldn't be marking you down since theoretically it it only 99% ionisation the first time. So writing equilibrium arrows is fine. That is what my chem teacher taught us anyway
 

SuchSmallHands

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I was under the impression that all acids disassociations are under equilibrium ? Markers I believe shouldn't be marking you down since theoretically it it only 99% ionisation the first time. So writing equilibrium arrows is fine. That is what my chem teacher taught us anyway
Yeah it should be at an equilibrium heavily to the right. Putting the equilibrium arrow in is technically more correct generally, though if this were a 2 marker that required you to write the equation you would put a normal arrow in as the question here specifies to assume complete ionisation.
 

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