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    what do i need to do to work out ph change ?

    i have worked out the ph from the strong base and strong acid but do not know what to compare them to?

    Q1: Calculate the pH change to 100 cm3 of 0.200 mol dm-3 HCl solution in a flask if 50 cm3 of 0.100 mol dm-3 NaOH is added.

    Q2: Calculate the pH change to 50 cm3 of 0.150 mol dm-3 KOH solution in a flask if 50 cm3 of 0.100 mol dm-3 H2SO4 is added.
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    (Original post by esmeralda123)
    what do i need to do to work out ph change ?

    i have worked out the ph from the strong base and strong acid but do not know what to compare them to?

    Q1: Calculate the pH change to 100 cm3 of 0.200 mol dm-3 HCl solution in a flask if 50 cm3 of 0.100 mol dm-3 NaOH is added.

    Q2: Calculate the pH change to 50 cm3 of 0.150 mol dm-3 KOH solution in a flask if 50 cm3 of 0.100 mol dm-3 H2SO4 is added.
    Use the stoichiometry of the reaction to find out the moles of the excess reagent (don't forget that the volume changes) and then calculate pH or pOH as appropriate.
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    (Original post by charco)
    Use the stoichiometry of the reaction to find out the moles of the excess reagent (don't forget that the volume changes) and then calculate pH or pOH as appropriate.
    i have but what do i compare the calculated value to?
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    (Original post by esmeralda123)
    i have but what do i compare the calculated value to?
    There is no indication in your question of comparison.
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    (Original post by charco)
    There is no indication in your question of comparison.
    pH change
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    Hey I'm not completely sure im correct but it seems like to me that you should first calculate the number of moles of HCL and the number of moles of NaOH. Calculate the difference and that'll be you're new number of moles of HCL as the acid has reacted with the base. Than calculate the new concentration, making sure you use 150cm3 and not 100cm3, and calcuate the pH using that value (-log(conc.). Thats you're new pH and you're old pH you can calculate using you're original values.

    The same thing for the second question only making sure that when you calculate the new pH, you make sure you divide the ionic product of water by the conc.

    Hope that helped, ask if you still don't understand!
 
 
 
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