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Q8 2003 Hsc (1 Viewer)

KungPow

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This is probably really obvious but...

8) A sulfuric acid solution has a concentration of 5 × 10^−4 mol L−1.
What is the pH of this solution, assuming the acid is completely ionised?
(A) 3.0
(B) 3.3
(C) 3.6
(D) 4.0

I don't see why the answer is 3. Do I have to do something to the number?
 

DAAVE

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H2SO4 is diprotic

so times the H+ concentration by 2
 

jumb

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Because sulfuric acid releases 2 h+ for each mole

-Log(2 x 5 x10^-4) = 3

It took me a few minutes to work this one out too :)
 

~*HSC 4 life*~

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when H2S04 is ionised it goes like this

H2SO4--> 2H+ + SO4 (2-)

note the 2 H+means u gotta double it
 

Paroissien

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H+ concentration is 2 times the sulfuric acid concentration as it is diprotic
Therefore, [H+] = 1 * 10^-3
pH = -log[H+]
= 3

EDIT: Wow people are answering questions quickly at the moment
 

DAAVE

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~*HSC 4 life*~ said:
when H2S04 is ionised it goes like this

H2SO4--> 2H+ + SO4 (2-)

note the 2 H+means u gotta double it
Technically it's:

H2SO4 + H2O <---> HSO4(-) + H3O(+)

and

HSO4(-) + H2O <----> SO4(2-) + H3O(+)

aint it?
 

Tommy_Lamp

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What everyone has said so far, since Sulfuric Acid is H2SO4, it is diprotic, that means there is double the concentration of hydrogen ions, hence you must multiply the concentration by 2, then determine the pH.
H+ = (5x10^-4) x 2
= 1x^-3
- log (1x^-3) = 3
therefore pH = 3
 

KungPow

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Wow what an explosion of help! Looks like everyone is eager to practice eh?

Well thanks for the help everyone. Damn diprotic. Always gets me!
 

Paroissien

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If it has a two next to the H it is diprotic
H2SO4 ... 2 next to the H
 

Tommy_Lamp

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smallcattle said:
how to find out when something is diprotic???
You need to know the chemical formula of the acid. For example:
HCl (Hydrochloric Acid) is monoprotic
H2SO4 (Sulfuric Acid) is diprotic
H3PO4 (Phosphoric Acid) is triprotic
 

Heinz

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DAAVE said:
Technically it's:

H2SO4 + H2O <---> HSO4(-) + H3O(+)

and

HSO4(-) + H2O <----> SO4(2-) + H3O(+)

aint it?
Sulfuric acid is still a strong acid, so its first ionisation goes to completion. The second stage is in equilibrium

H<sub>2</sub>SO<sub>4</sub> + H<sub>2</sub>O ==> HSO<sub>4</sub><sup>-</sup> + H<sub>3</sub>O<sup>+</sup>

HSO<sub>4</sub><sup>-</sup> + H<sub>2</sub>O <==> SO<sub>4</sub><sup>-2</sup> + H<sub>3</sub>O<sup>+</sup>
 

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