• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

eeek (1 Viewer)

Maddy1012

Member
Joined
May 9, 2006
Messages
79
Gender
Female
HSC
2007
In this dot point book for chem it has questions like

What is the pH for 0.01 of Sulfuric acid and its answer is 0.7... i thought it would be ph of 2.
When you apply -log [h+] it doesnt work.
Would they ask questions like that in the exam or will they always give u questions where you just apply -log [h+]

anyone help me

and how do you do ones where its like

calculate the pH when only 8% is ionised?
 

Martyno1

oh hi
Joined
Feb 24, 2006
Messages
762
Location
Sydney
Gender
Male
HSC
2007
Because sulphuric acid is diprotic, it has two ionisations, when you worked out the pH you only counted one. The theoretical pH would be -log 0.02, but I'm not sure if that is the actual pH of 0.01 mol/L sulphuric acid.

With the 8% thing, do you have an example of a question?
 

undalay

Active Member
Joined
Dec 14, 2006
Messages
1,002
Location
Ashfield
Gender
Male
HSC
2008
Maddy1012 said:
In this dot point book for chem it has questions like

What is the pH for 0.01 of Sulfuric acid and its answer is 0.7... i thought it would be ph of 2.
When you apply -log [h+] it doesnt work.
Would they ask questions like that in the exam or will they always give u questions where you just apply -log [h+]

anyone help me

and how do you do ones where its like

calculate the pH when only 8% is ionised?
0.7 or 1.7?
 

Mattamz

Member
Joined
Aug 13, 2005
Messages
64
Gender
Male
HSC
2007
for the 8% ionisation does anyone agree that it would be

pH = -log{[H+]*(8/100)}
 

jlnWind

Member
Joined
Sep 13, 2006
Messages
50
Gender
Male
HSC
2007
wrxsti said:
answers are clearly wrong....
Yeh its clear that the answers are wrong.
-log[0.02] = 1.69 (i.e. 1.7) they've mistyped it for sure

And its a bad question from the start because as marty said
its diprotic meaning there are two equations for the ionisation of H2SO4
i) H2SO4 + H2O --> H3O(+) + HSO4(-)
ii) HSO4- + H2O <--> H3O(+) + SO4(2-)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top