Question for anyone with Acer Sample Papers; unit 12 questions 36-38

[katie1980]

Anyone out there able to help with question 37... not sure how to approach it...


all help welcome.. thanks

[whenwhere]

Edit- Nevermind! Wrong unit, i'm looking at the wrong paper!

[katie1980]

hey.. thanks, for your help.. but yep wrong unit.. if anyone out there can help with the one I'm talking about.. acid/bases.. I'd be forever grateful....

[whenwhere]

Post the question. I can help with acid/bases.

[katie1980]

thanks whenwhere.. think I've figured it out.. it was just to do with interaction with indicators.. head is a bit wrecked with it all.. wish this blasted exam was tomorr

[terrified_tony]

you sorted?

[katie1980]

think so.. for now anyway.. thanks terrified tony...

[katie1980]

Question.. all help appreciated...

Pure water undergoes self-ionisation according to the equation H20-> H+ + OH-

The equilibrium constant for the rxn is as written 1x10^-14 at 25 degrees C and 1x10^-13 at 60 degrees.

At 60 degrees water has a pH value that is??

I know that the answer is less than 7 but the water is still neutral. But in the solutions is says;

"the self ionization of water will produce equal concentrations of H+ and OH- ions and so will always be neutral. ( this next bit is the bit I don't get...) At 60 degrees the concentration will be 3.2 x 10^-7 M and so the pH will be less than 7.

I don't understand or know where the 3.2 x 10^-7 has come from....

Anyway one help???

[terrified_tony]

been a while since i did this kinda thing, but believe the reason is that the equation for equilibrium constant is
equal to [H+] [HO]- / [h20] which due the the number of H+ and HO- being the same can be said to be
equal to H+ squared therefore H+ squared is equal to 1x10^-13 so squre rooting gives the answer....
3.2x10-7, Think thats where we're going with that one!

[katie1980]

staring me in the face.. thanks Tony... missed the square...