SHIFTED!!
Posted by shannontock at 06:33 PM on June 26, 2005.
cya!!
Posted by shannontock at 06:33 PM on June 26, 2005.
Posted by shannontock at 05:18 PM on June 26, 2005.
Posted by shannontock at 08:41 PM on June 24, 2005.
BOO HOO HOO !!!!!!!!!!!!!!!! i'm super sad!! 
Had my PIPC2 quiz at 5pm just now. Started at 508, ended at 548. KAOZZ so lil time. Actually is enough time one!! After i did my qn 2 then the tchr give the atomic mass, zzz! want to give give earlier la, lemme waste about 1min to redo!!! then after i finished the whole paper, i'm wondering, "eh, how come never make use of the Ka table provided ah?? and i nv do any dissociation of weak acids or bases??? WEIRD!" so i looked thru the qn, and .. =X !!!!! qn 1(b) i did wrongly!!!!!! realized HClO is a WEAK acid, but too late!!!!!!!!!!!!! CRY! didn't managed to finish tat qn in time, worth 5 or 6 marks loh!!! i know how to do one loh!! so wasted!!! veryvery very very very very very very very very SAD!!!!
the qn is : Calculate the pH of 0.5L of 0.0085M of HClO.
Initially, i dunno if HClO is Weak or strong acid lo. cuz the fucking HClO4 is a strong acid, then i did it in the STRONG acid method. then i looked thru the Ka table (which is only applicable to WEAK acids), i found HClO!!!!!!!!!! but it's tooo late! gimme 20sec more and i can finish the qn, and earn 6 marks!! booooooo hoooooooooooooo!
the ans is:
HClO (aq) ↔ H+ (aq) + ClO- (aq)
Initial: 0.0085 M 0 M 0 M
Change: -X +X +X
Eqlm: (0.0085-X) M X M X M
Let X = [H+]
Ka = [H+][ClO-] / [HClO], Given Ka = 3.0 X 10-8
3.0 X 10-8 = X2 / (0.0085-X)
Assume X<5% Concinitial, 0.0085 >> X,
thus, (0.0085-X) ≈ 0.0085M
X2 = 0.0085 x 3.0 x 10-8
X = 1.6 x 10-5
Test assumption, X<5% [HClO]initial
5% x 0.0085 = 4.2 x 10-5
Assumption valid.
So, [H+] = 1.6 X 10-5
pH = - log [H+] = - log (1.6 x 10-5)
= 4.80 !!!!!!!!!!!!!!!!!!!!!