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    I'm trying to do a unit 4 paper and I am so stuck on this question and it's only 2 marks can anyone help?!

    This question involves calculations about two strong acids and one weak acid.
    All measurements were carried out at 25 oC.
    4 (a) A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and
    100cm3 of distilled water were added.
    Calculate the pH of the new solution formed.
    Give your answer to 2 decimal places.
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    So far I have tried, and came up with :
    the conc of h30+ is the same as the acid but that's wrong. Oh no i'm really confused.
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    (Original post by Mariah246)
    I'm trying to do a unit 4 paper and I am so stuck on this question and it's only 2 marks can anyone help?!

    This question involves calculations about two strong acids and one weak acid.
    All measurements were carried out at 25 oC.
    4 (a) A 25.0 cm3 sample of 0.0850 mol dm–3 hydrochloric acid was placed in a beaker and
    100cm3 of distilled water were added.
    Calculate the pH of the new solution formed.
    Give your answer to 2 decimal places.
    Find the new concentration (hint: it's diluted by a factor of five) and then apply pH = -log[H+]
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    (Original post by charco)
    Find the new concentration (hint: it's diluted by a factor of five) and then apply pH = -log[H+]
    Is the original conc of h3o+ 0.0850
    and so you multiply that with 25/125 to get the new conc of h30+ (which is 0.017?
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    (Original post by Mariah246)
    Is the original conc of h3o+ 0.0850
    and so you multiply that with 25/125 to get the new conc of h30+ (which is 0.017?
    Ok, I don't know if your answer here is correct. But with this type of question, you need to find [H+]. So we use the volume and the conc given to find our molar value. Then, because the volume has increase, we find our new concentration by simply dividing are molar value over both the old and the new additional volume combined. This gives us our conc for our acid, because it is strong, you know what to do now with regard to finding the conc of H+
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    (Original post by mclovin123)
    Ok, I don't know if your answer here is correct. But with this type of question, you need to find [H+]. So we use the volume and the conc given to find our molar value. Then, because the volume has increase, we find our new concentration by simply dividing are molar value over both the old and the new additional volume combined. This gives us our conc for our acid, because it is strong, you know what to do now with regard to finding the conc of H+
    thanks, do you know which one or H+ and h30+ we are supposed to use in the exam ? My teacher said it's the same thing?
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    (Original post by Mariah246)
    thanks, do you know which one or H+ and h30+ we are supposed to use in the exam ? My teacher said it's the same thing?
    It is the same thing, it's just that H+ ions have a too large charge densities so they strongly attract water molecules which mean they are hardly ever on their own. We use H+ as it's just a simpler version.

    You should use H+ as this is what the Mark scheme will usually use and it makes things much easier.

    Spoiler:
    Show
    Don't want to confuse you any more but if asked to identify the acid-base pairs for the dissociation of HA in water for example, you would need to use H3O+ as it doesn't work otherwise e.g.
    HA + H2O <----> H3O^+ + A^-
    A1 + B2 <-------> A2 + B2



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    (Original post by Mariah246)
    Is the original conc of h3o+ 0.0850
    and so you multiply that with 25/125 to get the new conc of h30+ (which is 0.017?
    yes
 
 
 
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