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ocr a f325 revision thread

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or did you just make it up?

Not really, it can be anything. If that was fifty, then that could be used as well. The point is that I wanted to make it such that only 1 & 3 could be used so that reaction is eliminated. Or do you have to make them all make sense mathematically? Idk. Yep I did make it up.

Did I do it wrong? Uh-oh. That took time...and brain cells.

Reply 2621

DudeBoy

Original post by KD35

I dont think theyll ask us that as the 2nd dissociation is counted as a weak acid.

The spec says monoprotic acids only :smile:

Reply 2622

FailedMyExams

Help please,

7ai

http://ocr.org.uk/Images/61528-mark-scheme-unit-f325-equilibria-energetics-and-elements-june.pdf

Reply 2623

bmpink

Original post by kevloui

Ok let me warm you guys up..

Construct the equation for the oxidation of acidified iron(II) ions by oxygen

Fe2+ + o2 + 4H+ > Fe3+ + 2H2O ????

Reply 2624

kevloui

Original post by bmpink

Fe2+ + o2 + 4H+ > Fe3+ + 2H2O ????

niiiiiiceee; although its 4 moles of the iron, but nonetheless..
Describe a test to show the presence of iron(III) ions in a solution of FeSO4.7H2O.

Reply 2625

physicshelp123

Original post by kevloui

Ok let me warm you guys up..

Construct the equation for the oxidation of acidified iron(II) ions by oxygen

fe2+ + 3/2 O2 -> fe203 ??? or does there need to be H's

Reply 2626

Myocardium

Original post by bmpink

[ ] of reactants or products will increase or decrease to restore Kc to original value. It depends which side of the equation has the fewest moles as it will shift to the side with the fewest, due to more collisions per second. Kc only changes by temperature.

K, similarly, only temp. K will increase if temperature increases, more collisions per second.

By 'it will shift to restore Kc' your talking about the equilibrium of the equation right? :s-smilie:

And just to make sure, [ ] means conc. right?

Reply 2627

ofudge

Original post by bmpink

Someone please help me! I'm being stupid but for some reason I just can't see the answer.

The student adds 50cm3 of 0.250moldm-3 butanoic acid to 50cm3 of 0.05moldm-3 sodium hydroxide. Calculate the pH of this buffer. Ka = 1.51 x 10^-5 moldm-3. (2d.p.s)

June 2012 :smile:

For a Buffer, the Acid is always in Excess!

Acid: (50/1000) x 0.250 = 0.0125 moles
Base = 50/1000 x 0.05 = 0.0025 Moles

Therefore: the amount of Acid that has reacted = (0.0125-0.0025) = 0.01

Then using equation: [h+] = Ka Acid/Salt

[h+]1.51 x10-5 x (0.01/0.0025) = 6.04 x10-5

Finally: -log10(6.04x10-5) = 4.22 to 2 d.p

Reply 2628

kevloui

Original post by physicshelp123

fe2+ + 3/2 O2 -> fe203 ??? or does there need to be H's

There are H+'s somewhere

Reply 2629

ofudge

Original post by physicshelp123

fe2+ + 3/2 O2 -> fe203 ??? or does there need to be H's

There needs to be hydrogens as your charges dont balance..!

Reply 2630

_HabibaH_

Original post by bmpink

Ok, firstly lets look at what we're given:

50cm3 of 0.250 moldm-3 butanoic acid
50cm3 of 0.0500 moldm-3 sodium hydroxide
Ka of butanoic acid is 1.51x10-5 moldm-3

Firstly we know that butanoic acid and sodium hydroxide have different quantities and thus one is in excess, thus we need to work out how much of each chemical reacts.

To do this you world out the moles for each chemical:

Concentration x Volume / 1000 = Moles.

So let's work out the moles of each substance to see which one is in excess:

For butanoic acid:

50 x 0.250 / 1000 = 0.0125 moles of butanoic acid

For Sodium Hydroxide:

50 x 0.0500 / 1000 = 2.5x10-3 moles of sodium hydroxide.

From these to mole values we can see that butanoic acid is in excess, thus we need to work out how many moles of it react with sodium hydroxide.

0.0125 reacts with 2.5x10-3 moles
0.0125 - 2.5x10-3 = 0.01 moles of butanoic acid that has reacted with ALL of the NaOH.

Now that we have the moles of Butanoic acid and Sodium Hydroxide we can finally work out the answer in a few more steps:

First let's look at the PH buffer formula for reference:

H+ = Ka x [HA (moldm-3)] / [A- (moldm-3)]

Keep in mind that we need the values in concentration, yet we have moles. So we convert Moles to concentration.

There is only one difference though, in the buffer solution the volume of both chemicals are added together, thus the volume is actually 100cm3

So the concentration of Butanoic acid in the buffer solution is:

0.01 x 1000 / 100 = 0.1moldm-3 of Butanoic acid in the buffer solution.

And NaOH: 2.5x10-3 x 1000 / 100 = 0.025 moldm-3 of NaOH.

So we enter these values into the equation:

H+ = 1.51x10-5 moldm-3 x [0.1 (moldm-3)] / [0.025 (moldm-3)]

H+ = 6.04x10-5 moldm-3

pH = -log(H+)

pH = 4.2189

pH = 4.22

That took time to do, hopefully you get it.

I WOULD LIKE TO STRESS THAT THIS IS NOT MY WORK BUT KINDLY EXPLAINED BY JAKES123 - CREDIT TO THE USER. IT EXPLAINS IT REALLY WELL!

Reply 2631

kuku2013

Original post by kevloui

niiiiiiceee; although its 4 moles of the iron, but nonetheless..
Describe a test to show the presence of iron(III) ions in a solution of FeSO4.7H2O.

is this in the spec

Reply 2632

kevloui

Original post by kuku2013

is this in the spec

LOL yea it is, its to do with ligand substitution

Reply 2633

MedMed12

Original post by _HabibaH_

Yep i got C in both. I went down for both in Jan. I'm half a year wiser now...I hope!

what do you mean went down? had you sat the paper before this Jan?
hopefully we'll be OK. Guessing you need an A for chemchem?

Reply 2634

bmpink

Original post by Myocardium

By 'it will shift to restore Kc' your talking about the equilibrium of the equation right? :s-smilie:

And just to make sure, [ ] means conc. right?

The position of equilibrium changes because the concentrations of the reactants change, because if there's more on one side and less on the other... ! Sorry yes it does, force of habit! My teacher ALWAYS wrote it like that haha.

Reply 2635

otrivine

Original post by reneetaylor

The way I looked at it was like this:
When I doubles in expt 2 and 2 so does the initial rate OR when it quadruples in 1 and 3, so does the initial rate, when the H doubles in 1 and 3, the initial rate does not double. That's how I know it's definitely zero order.
Due to the proportionality of the orders with the initial rate, I think I is 1st order

(I'm not sure this is the correct way to do this, but it works for me every time)

I hope this helped a little!