Calculating the pH of acid, base AND WATER

The water doesn't interfere with the acid or alkali, it only increases the total volume of the solution, thus diluting it making the solution less concentrated.

So as a general rule here, you calculate which is in excess, the acid or the alkali in moles. Then you calculate the concentration of [H+] from this, taking into the account the increase in total volume due to adding 55ml of water. Then we can find PH.

So to begin:
Moles of H+ in H2SO4=CxVx2= 0.001x(25x10^-3)x2=0.00005moles of H+
Moles of OH- in NaOH=CxV= 0.001x(20x10^-3)= 0.00002moles of OH-

Moles of H+ in excess = 0.00005-0.00002=0.00003

Where V=25+20+55=100ml
Therefore, [H+]=Mol/V= 0.00003/(100x10^-3) = 0.0003moldm^-3

So, PH=-log[H+]
PH=-log(0.0003)
PH= 3.52 (2dp)

Any chance you could help me with this, please?

I'm having a hard time with these


Calculate how many ml of 0.2M KH2PO4 solution should be mixed with 0.1M K2HPO4 in order to obtain a buffer pH of 7.7 (pKa of H2PO4 = 7.41)

We don't learn that in AQA o_0
I guess that explains a lot.

It's not, the only equation's we know that's similar to that is that in AQA is:

PH=pKa

But not the +(log[A-]/[AH] part onto it. That equation isn't in the syllabus and has never come up in AQA exams.

Henderson–Hasselbalch equation:

pH = pKa + log([A-]/[AH])

In your example
7.7= 7.41 + log ([K2HPO4]/[KH2PO4])
rearrange
0.29= log ([K2HPO4]/[KH2PO4])
10^(0.29)= [K2HPO4]/[KH2PO4]
1.95= the ratio of [K2HPO4]/[KH2PO4]

Hope this helps.

Try using the Henderson-Hasselbach equation.

But you know

$K_a = \frac {[H^+][A^-]}{[HA]}$

That's the same equation, just rearranged. And it can be used to solve this problem, using Henderson-Haselbalch equation is just an equivalent approach, often faster.

This "this equation is not in syllabus" approach will make you fail at some point. Not my problem though.

Any chance you could help me with this, please?

I'm having a hard time with these


Calculate how many ml of 0.2M KH2PO4 solution should be mixed with 0.1M K2HPO4 in order to obtain a buffer pH of 7.7 (pKa of H2PO4 = 7.41)

But you know

$K_a = \frac {[H^+][A^-]}{[HA]}$

That's the same equation, just rearranged. And it can be used to solve this problem, using Henderson-Haselbalch equation is just an equivalent approach, often faster.

This "this equation is not in syllabus" approach will make you fail at some point. Not my problem though.

Hi,

I'm ok calculating the pH's when its just acids and bases - but when you're adding water too, what do you do?

e.g.

1) Calculate the pH of a solution prepared by mixing 25ml of 0.001 mol/l of H2SO4 and 20ml of 0.001 mol/l NaOH and 55 ml of water.


How would you go about doing this?

I'd greatly appreciate any help!

Just attempting another question:

A buffer was prepared by mixing:

100 ml of 0.1 mol/l Acetic acid
70ml of 0.1 mol/l NaOH

Calculate the pH.

I'm getting for my answer 5.07.

Can anyone kindly quickly confirm?

what's the ka of acetic acid?

Sorry, I should have mentioned it.

The pka is 4.7

So ka = 1.995 x 10-5