btw we haven't started this topic so im unfamiliar with these constants
It's 22.4 at standard temperature and pressure but 24.0 at room temperature and pressure. Are you in the UK because you should have done this by now? So go with the 24.0. What's the answer?
It's 22.4 at standard temperature and pressure but 24.0 at room temperature and pressure. Are you in the UK because you should have done this by now? So go with the 24.0. What's the answer?
I wouldn't say purple, the end-point is the first hint of a pink tinge colour.
The reason why it's colourless to pink is because exactly what you said. If you add Mn7+ to Fe2+, they will be reduced to Mn2+ which is colourless.
As I keep adding and adding the Mn7+, all the Fe2+ will eventually become oxidised to Fe3+.
There is a single moment when I have fully oxidised all the Fe2+ in the conical flask, and so at that moment, there is no more Fe2+.
Thus, the instant I add more Mn7+, it's going to remain as Mn7+ because there's nothing to oxidise. Thus we get this pink tinge colour. Add some more and we get that intense purple colour.
Unless you wanted to use some sort of indicator which changes colour around the neutralisation point, you would look for this colour change. And we tend not to because it is a prominent change.
Trying to find some people who can help me out with an F335 OCR Chemistry question - don't know if you're able to help! (No worries if not!)
I have managed to get the first two questions: pH = 3.33 & volume = 4.5 However I cannot for the life of me manage the last question! Thank you in advance
Trying to find some people who can help me out with an F335 OCR Chemistry question - don't know if you're able to help! (No worries if not!)
I have managed to get the first two questions: pH = 3.33 & volume = 4.5 However I cannot for the life of me manage the last question! Thank you in advance
There's a REALLY LONG way, and there's a much shorter way.
Here's the short way, because **** the long way. Lol I say short, but my post is long. That's only because I just really really want you to understand it sorry. The working out when you come to write it is short.
You agree that you have some moles of HA. The question is saying that the student adds some NaOH. This amount is equal to a third of the amount needed to neutralise it.
In other words, a third of the HA has reacted with the NaOH. If it said, half the amount was added, then half of the HA would have been neutralised, and so on.
When we react the HA and NaOH, this forms our A-.
Spoiler
So as a result, the amount of A- we formed is equal to one third of the HA we started with.
In other words...
A−=31×HA.
If you want to represent this as concentration, by all means divide it by the volume. But the volume will cancel out, which allows me to just use the moles.
In fact I don't need to do that because...
H+=A−Kc×HA becomes H+=31×HAKc×HA
And now the HAs cancel out, which just leaves...
H+=Kc×3
This is now solvable for pH It's just like that first question of where the HA is half the number of A
Can someone please explain the meaning of the Latin numbers next to ions (e.g Sulphate(IV)) and what effect they have on the number of oxygen atoms?
The latin numerals show the oxidation states. For example, sulphate (iv) means that sulphur has a oxidation state of +4. But we see ''ate'' which shows there is an oxygen but we don't know how many.
So if sulphur is +4 then there must be two oxygen. Because each oxygen is -2 and -2*2 is -4. So if we have sulphur 4+ and 2 oxygen being 4- (remember there is two oxygens each being -2 and -2*2 is -4) the charges will balance to give zero. Therefore the formula for this compound is SO2 Hope I helped you!