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    "50cm3 of 0.1mol dm-3 solution of sulphuric acid is to be neutralised by KOH as indicated by the equation below

    H2SO4 + 2KOH → K2SO4 + 2H2O

    what volume of 0.05 moldm-3 KOH will be required?"
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    Sorry you've not had any responses about this. Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses.

    I'm going to quote in Tank Girl now so she can move your thread to the right place if it's needed. :yy:

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    (Original post by tomlfc)
    "50cm3 of 0.1mol dm-3 solution of sulphuric acid is to be neutralised by KOH as indicated by the equation below

    H2SO4 + 2KOH → K2SO4 + 2H2O

    what volume of 0.05 moldm-3 KOH will be required?"

    Number of moles = concentration x volume

    You know the concentration and the volume of sulfuric acid to be 0.1mol dm^-3 and 50cm^3.
    first you need to convert 50cm^3 into dm^3.
    Then you can work out the number of moles in sulfuric acid.
    Using that, you're able to find the number of moles in potassium hydroxide as you know the ratio is 1:2
    You're also given the concentration of potassium hydroxide so using the equation above, when rearranged, you can work out the volume

    Hope that kinda helped
 
 
 
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