pH Calculation Question [HELP!] (1 Viewer)

[mclavin]

Calculate the pH of the solution resulting from mixing:
5.0mL of 0.100M H2SO4 and 25.0mL of 0.100M NaOH.

I don't know the answer but I'm pretty sure I've gone wrong somewhere, this is what I did though:

H2SO4 + NaOH --> 2H2O + Na2SO4

n(H2SO4) = C x V = 0.1 x 0.05 = 0.005 Moles
n(NaOH) = C x V = 0.1 x 0.25 = 0.025 Moles

And then I did Moles total of H+ = 0.025 - 0.005 = 0.020 Moles
Volume Total = 0.005 + 0.025 = 0.03 L

Therefore Concentration = n / V = 0.02 / 0.03 = 2/3
Hence, pH = -log(2/3) = 0.176

Was I wrong not to use pOH = -log[OH-] instead? I don't really understand this too much so thanks for any help!
Much appreciated.

 

[Librah]

Someone correct me if i made an error. And if you can't understand my handwriting lol, kind of rushed through it. Forgot units on 4th line. There's some s.f errors, but i cbb editing. Also best to write down the unrounded answer first then give the actual one.

 

[mclavin]

Still a bit unsure about the calculations of H+ and OH-, although I think I see where all the values come from.
Thanks very much though, that clears it up a bit.

 

[Librah]

Still a bit unsure about the calculations of H+ and OH-, although I think I see where all the values come from.
Thanks very much though, that clears it up a bit.

Which part?

 

[zhertec]

Yep, YunLi should be correct (got the same answer as him :3)

To OP - For these questions:
1. Write the balanced Equations of which the compounds dissociate/ionise in to H+/OH- ions.
2. Find the moles of H+ and OH-.
3. Find the Remaining moles of ions present after neutralisation. (just minus the total amount since everything is 1:1:1 ratio.)
4. Divide the moles of ions over the total volume after mixture to find the concentration (mol/L)

For H+:
-log (base 10) (Concentration of H+) for the pH.

For OH-:
Use the ionisation constant of water to find the amount of H+ ions: Since (H+)(OH-) = 10^-14 sub the known values in (basically 10^-14/concentration of OH- ions).
Then -log (base 10) (Concentration of H+ found from that step) for the pH.

 

[siggy]

Yep, YunLi should be correct (got the same answer as him :3)

2. Find the moles of H+ and OH-.
3. Find the Remaining moles of ions present after neutralisation. (just minus the total amount since everything is 1:1:1 ratio.)

This is the bit I'm still a bit unsure about. I can easily find the moles of H2SO4 or NaOH and can make the equation but I don't entirely understand your working to find the number of moles of H+ and OH-. And then I don't really understand why / what's going on when we are finding the remaining ions? I assume we minus because one is neutralising the other or something...?

Thanks for the help both of you!

 

[Librah]

This is the bit I'm still a bit unsure about. I can easily find the moles of H2SO4 or NaOH and can make the equation but I don't entirely understand your working to find the number of moles of H+ and OH-. And then I don't really understand why / what's going on when we are finding the remaining ions? I assume we minus because one is neutralising the other or something...?

Thanks for the help both of you!

You could say I'st not the actual acids themselves that are reacting with each other in Arrhenius terms. But it's the H+ and OH- ions that are "released" by the respective acid/base. Since the base is in excess, it will completely neutralise the acid. For every mole of NaOH, one mole of OH- is released, for every mole of H2SO4, we assume it undergoes complete ionisation and releases 2 moles of H+ ions. Then OH- is left in excess floating around. The remaining ions will determine the pH of the solution by definition of what pH is.

 

[mclavin]

It not the actual acids themselves that are reacting with each other. It's the H+ and OH- ions that are released by the respective acid/base. Since the base is in excess, it will completely neutralise the acid. For every mole of NaOH, one mole of OH- is released, for every mole of H2SO4, we assume it undergoes complete ionisation and releases 2 moles of H+ ions. Then OH- is left in excess floating around. The remaining ions will determine the pH of the solution by definition of what pH is.

Oh ok that's really helpful. Does H2SO4 release 2 moles of H+ because of its' valency?

 

[InteGrand]

Oh ok that's really helpful. Does H2SO4 release 2 moles of H+ because of its' valency?

No, it's because it's a strong acid.

Technically, only the first step of its ionisation undergoes complete ionisation: H₂SO₄ → H⁺ + HSO₄^-. (H₂SO₄ has two steps of ionisation.)

The second step is the ionisation of bisulfate (HSO₄^-): HSO₄^- ⇌ H⁺ + SO₄^2-. This last step technically doesn't undergo completion (I think about 80% completion is achieved), so 1 mole of H₂SO₄ will actually produce around 1.8 moles of H⁺ ions.

But for HSC purposes, we assume this second step goes to completion and that 1 mole of H₂SO₄ produces 2 moles of H⁺ ions.

 

[InteGrand]

There are acids with H⁺ ions but that don't produce H⁺ ions per mole. An example of this is phosphoric acid (H₃PO₄). Its first stage of ionisation goes to completion if I recall correctly, but its subsequent two don't, so you get somewhere between 1 and 3 H⁺ ions per mole of H₃PO₄ dissolved.

The acid dissociation constant K_a for an acid provides a measure of its strength, but this is not in the HSC syllabus. Basically, if for an acid's ionisation step, then it'll ionise to completion in that step. Otherwise, we don't have complete ionisation. K_a values for acids can be looked up in data tables. Bigger K_a values correspond to stronger acids.

 

[siggy]

So essentially H2SO4 fully ionises because it is a strong acid and the reason it ionises into 2H+ instead of one is because it is H2SO4?

 

[InteGrand]

So essentially H2SO4 fully ionises because it is a strong acid and the reason it ionises into 2H+ instead of one is because it is H2SO4?

For HSC intents and purposes, yes.

 

[DepressedPenguino]

Is the pH of the solution 2.7 or 12.7?

 

[DepressedPenguino]

I am assuming 12.7