Confusing Question

[Christiano]

In a 0.001 M solution of the weak acid C2H5COOH the pH will be closest to:
A:2
B:3
C:4
D:9

The expected answer is 3. However the actual answer is 4. Can anyone tell me why? I remember someone telling me the reason why, but I forgot :S

[UncleXxx]

sorry i don't know how to use latex.

Ka=1.3*10^-5)on calc and u shud get  a ph value of 3.943
Rounding to one significant figure will give you 4.

[Christiano]

Isn't it:

pH=-log(10 to the power of -3)
pH= 3?

Would the reasoning behind it be because it is a weak acid, the answer can be expected to be the closest number to 7 (neutrality)

[iffets12345]

Isn't it:

pH=-log(10 to the power of -3)
pH= 3?

Would the reasoning behind it be because it is a weak acid, the answer can be expected to be the closest number to 7 (neutrality)

You are kind of,  but not totally, on the right track.
It is a weak acid, therefore you can't expect complete ionisation of the CH3COO- and the H+ ions, unlike in HCL where they would almost totally ionise. Therefore, you won't have exactly 0.001 or whatever of the H+ ionising, it's going to be less than this and hence you won't have that precise pH of -log10(0.001)

[matt123]

they dont provide you with a KA .. and they dont give enough relevant data to calculate it.

sorry i don't know how to use latex.

Ka=1.3*10^-5)on calc and u shud get  a ph value of 3.943
Rounding to one significant figure will give you 4.

they dont give u the KA in the question.
i havnt looked . maybe its in the data booklet.

however
my understand is
even though the ph = 3 ( theoretically)
practically it shouldnt , and possibly = 4 ...
as its a weak acid.
it cannot ionise completely . therefore .. it has a higher ph than expected.

hmmmm how did u find the KA? did u use data book?

[UncleXxx]

they dont provide you with a KA .. and they dont give enough relevant data to calculate it.

sorry i don't know how to use latex.

Ka=1.3*10^-5)on calc and u shud get  a ph value of 3.943
Rounding to one significant figure will give you 4.

they dont give u the KA in the question.
i havnt looked . maybe its in the data booklet.

however
my understand is
even though the ph = 3 ( theoretically)
practically it shouldnt , and possibly = 4 ...
as its a weak acid.
it cannot ionise completely . therefore .. it has a higher ph than expected.

hmmmm how did u find the KA? did u use data book?

Last page of the data book

[matt123]

they dont provide you with a KA .. and they dont give enough relevant data to calculate it.

sorry i don't know how to use latex.

Ka=1.3*10^-5)on calc and u shud get  a ph value of 3.943
Rounding to one significant figure will give you 4.

they dont give u the KA in the question.
i havnt looked . maybe its in the data booklet.

however
my understand is
even though the ph = 3 ( theoretically)
practically it shouldnt , and possibly = 4 ...
as its a weak acid.
it cannot ionise completely . therefore .. it has a higher ph than expected.

hmmmm how did u find the KA? did u use data book?

Last page of the data book

haha yeh i didnt look
i guess its a tricky question
many people would jump into it .. and ignore the fact u need 2 use the KA value since its a weak acid.
good question.

[simonhu81292]

yep ...last page on the data book ...
if this is unit 4 ..
possibility using that methods... equilibrium is introduced..
so i don't see why not

[iffets12345]

Did you guys go through acid -base equilibrium? my teacher made it explicit from the start that all those KA values would be in the booklet if you werent given them any other way....that's very important =/.

[shinny]

Yep, given that it's a multi choice and they say choose the best possible answer...
a. It cannot be 2 because if it was a strong acid, it would have a pH of 3. Hence, a weak acid of this concentration can't be pH 2 because it can't be more acidic than a strong acid.
b. Similar logic to above, but a weak acid can't have the same pH of a strong acid of the same concentration.
c. Still acidic, but less acidic than would be expected of a strong acid of the same concentration. Hence, this.
d. Can't be because it's basic. You can't add an acid and make it turn basic.

But yes, in the exam you'll have all the Ka values. It's faster to solve this question by deduction though.

[vea]

I thought that you need to find the concentration of just the H3O+ and not the concentration of the whole solution.

Can someone please explain how this is done?

[matt123]

Did you guys go through acid -base equilibrium? my teacher made it explicit from the start that all those KA values would be in the booklet if you werent given them any other way....that's very important =/.

nah i tottally forgot about the data book.
alot of people will . they will jump the question and try to solve it by " -log ( 0.001) instead of using the KA valuse and ( H3O+)^2 / 0.001 ....

very easy question .. but smart question that will throw a few off

in our sac the KA values were given . so i forgot bout data book .. but yeahhh never forgot again now haha
thanks guys

[matt123]

I thought that you need to find the concentration of just the H3O+ and not the concentration of the whole solution.

Can someone please explain how this is done?

u do .. to get the PH u always have to know the ( H30+) concentration

hence ..
if they tell u the concentration of a WEAK ACID is 0.001
u know the formula

KA = (h30^2) / weak acids concentration

therefore

KA (1.3X10^-5) = h30^ 2 / 0.001

rearange the formula
H30+^2 = 1.3X10^-5 X 0.001

then .... square root of that = 0.000114 which is the H30

-log of that = 3.8
round up
u get 4

goodluck
hope that helps buddy.

[iffets12345]

Just simply, if a solution is 0.001, that does not mean the [H30+] is the same. It could be partially ionised as I said before.

[matt123]

Just simply, if a solution is 0.001, that does not mean the [H30+] is the same. It could be partially ionised as I said before.

yeh hes exactly right
if it is A WEAK ACID ...
this is the case

however if it is strong
such as hcl/h2s04
you can take 0.001 as the h30+

if its weak
YOU MUST USE KA .. YOU CANNOT GET AROUND THAT.
goodluck buddy