[Efrem22]

Firstly I apologize if this topic is already up, I searched before posting but didn't find anything that helped me out.


It's probably the easiest pH question there is out there but for some reason I can't seem to wrap my head around it. I'm not looking for the answer to it. Just clear steps on how to answer the problem.


The problem is;

what is the pH of a 2.9L solution containing .0200 mols of HNO3?

its a multiple choice question the answers I can choose from are;

a)2.16 b).30 c) 14.30 or d) 13.70

What I've tried hasn't given me any of those answers.

I've tried -log(.0200) which gives me 1.69 (1.7) but clearly that isn't any of the answers above. I've even tried scientific notation only to get the same answer.

any help will be greatly appreciated. Thanks.

[FreeTheBee]

Hi,

What is the concentration of HNO3 in this case?

[AWK]

pH 2.16 is the answer for dilution of 1.0 L 0f 0.02 M HNO₃ to 2.9 L.

[Efrem22]

Okay, thanks. How exactly did you go about figuring that, or should I just review my dilution notes? Or maybe look at an "ice chart" or whatever they are called. (It is kind of an easy question, I just missed the class they went over it all, and now I'm trying to self teach it.)

[Borek]

Start just applying definition of concentration - you have 0.02 moles of HNO₃ in 2.9L of water.

Then just apply definition of pH.

[Efrem22]

I really feel like an idiot now aha. Thanks a lot for the help. I guess I was just reading into it too much.

[Efrem22]

I have another question. More so to check my math in another question.

A solution has a [H+] = 4.6*10^-3 M. What is the [OH-] in the solution?
Again a multiple choice question.
A) 4.6*10^-11
B) 4.6*10^-17
C) 2.2*10^-12
D) 1.0*10^-13

Just by looking at it I'm pretty sure its B.

the math I did was

[OH-]= [H+]/1.0*10^-14
[OH-] = 4.6*10^-3/1.0*10^-14
[OH-] = 4.6*10^-17

[sjb]

[OH-]= [H+]/1.0*10^-14

How have you got this equation?

[Efrem22]

See that's why I'm asking because I wasn't sure if I was even using the right equation.

I kind of changed it around to work in my favour.

in my notes I have

[OH-] = 1.0x10^-14/[H+]

but now looking over my nots I think I see where I maybe went wrong. - what i tried also didn't work so back to square one.

I'm going to try something else. But all I ask is for a little direction, not the answer.

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

[chenbot]

I do it the long way, but it works every time and I don't get confused.

I do -log([H+]) then subtract that from 14, then do 10^(-whatever I get)