pH calculation

[jcpmed]

Hi, can anyone help?

I'm stuck with this question:

Calculate the pH for a solution prepared by mixing 25mL of 0.02M of HNO(aq) with 50mL of 0.01M HCL(aq) ?

I spent some time on it but can get the workings to get the answer pH =1.88, well I think thats the answer.

Thanks

jcpmed

[Sir_Dispensalot]

show us how you got to your answer (ie your workings), and we can see if it matches up with our calculations too.

[jcpmed]

Playing hard ball?

I have to prove my workings - thanks dad,

ok

PH = -log10 [h30+]

[h30+] = 0.2m plus 0.8M = 1M

PH = 1

or 0.02 plus 0.01/75 x 10-power 3 = 13.33

PH = 14-13.33 = .66 ??

a bit lost

its ok if you cant help

so rather stuck

[Sir_Dispensalot]

Playing hard ball?

I have to prove my workings - thanks dad,

ok

Not playing hard ball, but we aren't a 'do your homework' service either. Sorry to do this but far too many pharmacy students are simply trying to use us as an easy way to do their assignments. We're more than happy to help you along but we refuse point blank to do your degree for you so you can spend more time down in the union bar.

And it's Stepdad, not dad.

[Nik]

This is how (I think) you answer it:

25ml of 0.02M HNO3 = 25/1000 x 0.02M = 5x10^-4 moles.

50ml of 0.01M HCL = 50/1000 x 0.01M = 5x10^-4 moles.

[H+] = (5x10^-4) + (5x10^-4) = 1x10^-3 moles/litre.

ph = -log[H] = 3

[jcpmed]

Hi Nikolai

Thanks for your help. My tutor in her ans to this question said it was pH = 1.88?

You solution is logical. I have emailed her with the query, hopefully 3 is the right answer.

Thanks a million.

Im teetotal.

Best wishes !

[Nik]

If you could get her to explain how she got 1.88 it would be helpful so i could understand as well. PH 3 seemed logical to me because you have strong acids but molar concentrations are low (0.02M and 0.01M) so wouldn't have expected a lower value like 1.88

[jcpmed]

I have emailed her - so will let you know asap. I spent an hour plus on it myself. Im wondering if we need the Ka of HCL or HNO3?

Do we use this formula?

pH = pka plus log Salt/acid? to find the pH?

[Nik]

But you are adding to strong acids together, where do salts come into it ?

[jcpmed]

One is an acid and the other a conjugate base? thus forming a salt? its a thought?