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    Need help with this question.

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

    Not entirely sure how to do this question.

    The normal way I would workout this question if it didn't mention pH change is:

    mol H+ = (100)/(1000) *0.2 = 0.02

    mol OH- = 50/1000 * 0.1= 0.005

    Xs Mol H+ = 0.02- 0.005 = 0.015

    Mol H+ = 0.015/ (150/1000) =0.1

    pH= -log (0.1) = 1.

    Not sure of what the change is though???
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    (Original post by Super199)
    Need help with this question.

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

    Not entirely sure how to do this question.

    The normal way I would workout this question if it didn't mention pH change is:

    mol H+ = (100)/(1000) *0.2 = 0.02

    mol OH- = 50/1000 * 0.1= 0.005

    Xs Mol H+ = 0.02- 0.005 = 0.015

    Mol H+ = 0.015/ (150/1000) =0.1

    pH= -log (0.1) = 1.

    Not sure of what the change is though???
    pH comes from concentration of H+ ions

    Calculate the concentration initially ====> hence the pH

    Calculate the concentration after some has been neutralised ====> hence the new pH
 
 
 
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