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    Hello. I'm not an a-level chemistry student but I need to know about preparing solutions of varying pHs which I understand it taught in a-level Chemistry. I was hoping if perhaps one of you could be so kind as to please give me some guidance on something which I require for my biology A2 coursework. I understand that you're all busy with revision and coursework yourself but perhaps explaining to me will aid in your revision as well as be an opportunity to apply your knowledge of pH calculations. I would really appreciate your help as I've tried reading various online guides on this matter and I honestly can't comprehend it - I don't know how you Chemistry students do it!

    Anyway, enough of my rambling. Here is my issue:
    I need to prepare 5 different solutions of a volume of 30cm3. Each solution needs to have a different pH. The pH values I have chosen are 5, 6, 7, 8 and 9. I understand for pH 7 I can just use distilled water but is the remaining four pH values I am having trouble with.

    Thank you so much for even taking the time to read this thread. I hope you'll be so kind as to help me out here. I would really, really appreciate it.

    Thanks again.
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    Your facility should have pH buffer tablets that dissolve in water (although not easily) to give a given pH. But make sure it is deionised/distilled water


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    Well I'm not really sure if they do this for A level chemistry. I don't remember seeing this back then, it's from my 1st year in university.

    Acidity is caused by hydrogen ions (protons) in the solution. The pH is given by -log (H+) where H+ is the concentration of hydrogen ions in the solution. So the pH is essentially a measure of the concentration of hydrogen ions in a solution. So to make the solution to the required pH you would need to work out the concentration of hydrogen ions you needed initially. Concentration = number of moles/volume. So rearrange pH=-log(number of H+ions/volume of solvent) to -volume*10pH=number of H+ions I think.

    Strong acids completely dissociate so if you used a strong acid you would be able to gurantee that all of the hydrogen atoms in the molecules dissociated to give hydrogen ions. For example HCl contains 1 hydrogen atom, therefore 1 mole of HCl would dissociate giving you 1 mole of hydrogen ions.

    So once you know the concentration of hydrogen ions you need from the initial part make up a solution of that concentrion of acid.

    These are random made up values, but you find for pH X you need a concentration of hydrogen ions of 3 moles per litre. You are using Hydrochloric acid to make the required concentration which you know completely dissociates because it's a strong acid. So you know 3 moles of HCl will give you 3 moles of hydrogen ions in solution. So you dissolve 3 moles of HCl in the required solvent.

    Or for another example, you are using Sulphuric acid which you know completely dissociates but has the formula H2SO4. As it has 2 hydrogens which completely dissociate you know 1 mole of H2SO4 gives 2 moles of hydrogen ions so to make your solution of pH X instead of using 3 moles as you used for the HCl you would use 1.5 moles. (Of course remembering never to dilute concentrated Sulphuric acid solutions with water.)

    I hope this helps, though personally I think this might be a bit advanced.
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    I miss the days the only things I used to buff were my boots...
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    (Original post by darreng16)
    Your facility should have pH buffer tablets that dissolve in water (although not easily) to give a given pH. But make sure it is deionised/distilled water


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    Unfortunately we don't. I'm required to give in a list of the volumes of acid/bases I need by Thursday.
    (Original post by 19alex91)
    Well I'm not really sure if they do this for A level chemistry. I don't remember seeing this back then, it's from my 1st year in university.

    Acidity is caused by hydrogen ions (protons) in the solution. The pH is given by -log (H+) where H+ is the concentration of hydrogen ions in the solution. So the pH is essentially a measure of the concentration of hydrogen ions in a solution. So to make the solution to the required pH you would need to work out the concentration of hydrogen ions you needed initially. Concentration = number of moles/volume. So rearrange pH=-log(number of H+ions/volume of solvent) to -volume*10pH=number of H+ions I think.

    Strong acids completely dissociate so if you used a strong acid you would be able to gurantee that all of the hydrogen atoms in the molecules dissociated to give hydrogen ions. For example HCl contains 1 hydrogen atom, therefore 1 mole of HCl would dissociate giving you 1 mole of hydrogen ions.

    So once you know the concentration of hydrogen ions you need from the initial part make up a solution of that concentrion of acid.

    These are random made up values, but you find for pH X you need a concentration of hydrogen ions of 3 moles per litre. You are using Hydrochloric acid to make the required concentration which you know completely dissociates because it's a strong acid. So you know 3 moles of HCl will give you 3 moles of hydrogen ions in solution. So you dissolve 3 moles of HCl in the required solvent.

    Or for another example, you are using Sulphuric acid which you know completely dissociates but has the formula H2SO4. As it has 2 hydrogens which completely dissociate you know 1 mole of H2SO4 gives 2 moles of hydrogen ions so to make your solution of pH X instead of using 3 moles as you used for the HCl you would use 1.5 moles. (Of course remembering never to dilute concentrated Sulphuric acid solutions with water.)

    I hope this helps, though personally I think this might be a bit advanced.
    Thank you so much for posting that. It was easy to follow and I think I've managed to come up with some working out. Unfortunately it seems to be a tiny amount of acid which would be pretty impossible to measure.

    Conc of HCl = 10-5
    Conc of HCl needed = 1x10-5moldm-3
    Moles of HCl = conc X vol = (30x10-3)(1x10-5) (the 30x10-3) is 30cm3 in dm3
    Moles of HCl = 3x10-7
    3x10^-7/0.1 (I assumed we’d have 0.1molar conc of HCl at college) = 3x10-6 which is 0.003cm3

    Is there any way around this? I've heard of weak acids which have a higher pH but they are affected by diltution. Do you know how I could work out the volume of a weak acid (I know we have ethanoic acid at college) to create 30cm3of the pHs 5, 6, 8 and 9? Thank you so much.

    (Original post by Jacob :))
    I miss the days the only things I used to buff were my boots...
    Hahaha. Me too, me too.
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    Meh, the joys of pH solutions, good old unit 4 and thankfully I don't have to do more of it :lol:
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    (Original post by lovex)
    Thank you so much for posting that. It was easy to follow and I think I've managed to come up with some working out. Unfortunately it seems to be a tiny amount of acid which would be pretty impossible to measure.

    Conc of HCl = 10-5
    Conc of HCl needed = 1x10-5moldm-3
    Moles of HCl = conc X vol = (30x10-3)(1x10-5) (the 30x10-3) is 30cm3 in dm3
    Moles of HCl = 3x10-7
    3x10^-7/0.1 (I assumed we’d have 0.1molar conc of HCl at college) = 3x10-6 which is 0.003cm3

    Is there any way around this? I've heard of weak acids which have a higher pH but they are affected by diltution. Do you know how I could work out the volume of a weak acid (I know we have ethanoic acid at college) to create 30cm3of the pHs 5, 6, 8 and 9? Thank you so much.
    Hmmm let me think, do you know the pH of the stock solution they provide? if you knew the pH of the stock solution you could work out how to dilute it to the required concentration.

    Work out the concentration of the pH of the stock solution they have. Then work out the concentration of the required pH and dilute it accordingly?

    Ermmm I don't know if there is a simpler way of doing this because I'm doing it through logic rather than using an equation from a text book. But if you have a solution of 1 mol dm-3 and you wanted to dilute it to a 0.5 mol dm-3 solution. You could use the wolume you have to work out how many moles were in there, so you would have 1 mole in 1 litre. You want a concentration of 0.5 moles per litre. Concentration is number of moles /volume so you could rearrange to get number of moles/concentration = volume. So for this care you know you have 1 mole and you know you want a concentration of 0.5 mol dm-3 so 1/.5=2. So you can see you need 2 litres of solution. You already have 1 litre so just add 1 litre of distilled water and it will make up the required concentration.

    Another example
    You have a solution of 5.5 moles per litre and you have 2 litres of solution. You want to make a concentration of 2.3 moles per litre
    Conc=no of moles/vol so conc x vol = no of moles 5.5 x 2 = 11 moles.

    conc = no of moles/volume so no of moles/conc = volume you want a conc of 2.3 moles per litre. 11/2.3=4.78 litres. You originally had 2 litres so 4.78-2 = 2.78 litres need to be added.


    If you find the pH of the stock solution and need any help working it out get back to me in here and if I see it I will try and give you a hand. Also you don't wanna just make up 30cm cubed, i'm assuming you're going to do repeats.
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    (Original post by 19alex91)
    Hmmm let me think, do you know the pH of the stock solution they provide? if you knew the pH of the stock solution you could work out how to dilute it to the required concentration.

    Work out the concentration of the pH of the stock solution they have. Then work out the concentration of the required pH and dilute it accordingly?

    Ermmm I don't know if there is a simpler way of doing this because I'm doing it through logic rather than using an equation from a text book. But if you have a solution of 1 mol dm-3 and you wanted to dilute it to a 0.5 mol dm-3 solution. You could use the wolume you have to work out how many moles were in there, so you would have 1 mole in 1 litre. You want a concentration of 0.5 moles per litre. Concentration is number of moles /volume so you could rearrange to get number of moles/concentration = volume. So for this care you know you have 1 mole and you know you want a concentration of 0.5 mol dm-3 so 1/.5=2. So you can see you need 2 litres of solution. You already have 1 litre so just add 1 litre of distilled water and it will make up the required concentration.

    Another example
    You have a solution of 5.5 moles per litre and you have 2 litres of solution. You want to make a concentration of 2.3 moles per litre
    Conc=no of moles/vol so conc x vol = no of moles 5.5 x 2 = 11 moles.

    conc = no of moles/volume so no of moles/conc = volume you want a conc of 2.3 moles per litre. 11/2.3=4.78 litres. You originally had 2 litres so 4.78-2 = 2.78 litres need to be added.


    If you find the pH of the stock solution and need any help working it out get back to me in here and if I see it I will try and give you a hand. Also you don't wanna just make up 30cm cubed, i'm assuming you're going to do repeats.
    Ooh thank you. It is just 30cm3 I need though as it's 10cm3 per experiment and then three repeats resulting in 30cm3. I think it's going to be impossible though because for example I need a pH of 5. 10^-5=1x10-5moldm-3. If the concentration of HCl is 0.1 I will need to dilute it 10000 times and when I'm only using 30cm3 it would mean I'd need to add 0.003cm3 of acid which is too small to measure effectively.

    (Original post by James A)
    Meh, the joys of pH solutions, good old unit 4 and thankfully I don't have to do more of it :lol:
    Haha, this biology coursework has cemented the fact that I never want to do a-level chemistry in the near future.
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    Hmmm you could try a weaker acid, instead of the concentration though you would need to use the activity which I think tells you how many molecules dissociate to give you hydrogen ions. I haven't learn this yet though so don't want to advise you about it. Sorry.
 
 
 
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