Help with chemistry problem for college

[elrond]

I know this is probably a long shot, but........I would be hugelly grateful if any of you good folks out there could help me with a chemistry problem which I am struggling with. I missed a couple of lectures on this bit and am absolutely stumped.

The problem is this:

Calculate the pH of the solution formed by mixing 10.0 cm3 of 0.1moldm-3 H2SO4 with 10cm3 of 0,125moldm-3 NaOH

A worked example would be worth its weight in gold.

fingers crossed someone out there knows what this means.

 

[Wayne]

Been a while since i did my A level chemistry. PM Lithril he is a chemist and a teacher. He will be able to help if he is not too busy.

I will try and find my old notes in the loft.

 

[leon-b]

ouch my head hurts lol
leon

 

[Brocktor]

i may research it and work it out tomorrow, i just hope no1 else does it first!

 

[tomtom]

I don't miss chemistry, y'know

Hope someone can help you out mate, I'm afraid I'm not qualified.

 

[Floyd Soul]

Is that exactly how the question is stated?

Ive been doing a very similar question this week for my OU course.

To find out the pH you'll need to figure out the hydrogen ion concentration of the mixture. Have you any notes with that info?

 

[elrond]

Thanks all for your replies to date.

So far.

I have balanced the equation:
H2SO4 + 2NaOH ----- Na2SO4 + 2H20
I am ok with working out the no of moles in the H2SO4 & 2NaOH
as i understand it
H2SO4 = 1x10-3 moles
NaOH = 2.5x10-3 moles
I presume that once all of the H2SO4 is used up then 1.5x10-3 moles of NaOH remain.
This is in a volume of 20cm3 of solution.
It is at this point that my head hurts ;-) and I dont know what to do next.
can anyone confim what i have done so far is correct and the next steps.

thanks again

 

[stevec]

check this out at wiki if i were you, but

pH = -log(conc Hions)

i seem to recall something about [H][OH]=1x10^-14

if you substitute into this it shoud be possible to calclate pH

bear in mind that its been 9-12 years since i did this sort of chemistry

give it a try and see if it works

all the best
steve

 

[Lithril]

Ok I think I'm write in this its been a while and I'll need to check when I get home and have the text books handy (been about 7 years since I've done this)

Work out the mols of each:

H2SO4:

10cm = 0.01dm
0.01 x 0.1 mol dm = 0.001 mols

BUT each mol of H2SO4 will give 2 mols of H+

therefore

we have 0.002 mols of H+

NaOH
10cm = 0.01 dm
0.01 x 0.125 mol dm = 0.00125 mols of -OH

0.002 - 0.00125 = 0.00075 mols H+ left in 20cm3 after neutralisation.

in 1 dm = 0.00075 / 0.02 = 0.0375 mol dm

For a strong acid that fully dissociates pH = - log [H+]

so -log 0.0375 = 1.426 = pH of solution

Hope that helps

Matt

 

[Boxy]

Beat me to it!

I second that!

 

[anthonyyy]

-log(2*10*10^-3dm3 *.1*2moldm-3)- (10*10^-3dm3*.125moldm-3)

bloody students!

 

[elrond]

Yet again, you folks leave me stuck for words.

you are all a great bunch and it really is a priviledge to share interests with you.
I just hope that at some point I can do something to help you.

Thanks again :You_Rock_

Gary