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    At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value 1.75 × 10–5 mol dm–3.
    (c) (i) Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassiumhydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C.(4 marks)
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    Name:  IMG_20160612_214744.jpg
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Size:  318.2 KB not sure but thats how i would do it. Is answer wrong or right?
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    (Original post by chhhhelsie)
    Name:  IMG_20160612_214744.jpg
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Size:  318.2 KB not sure but thats how i would do it. Is answer wrong or right?
    It's much easier than that.

    After reaction moles of acid = moles of salt

    Therefore pH = pKa = 4.75
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    (Original post by charco)
    It's much easier than that.

    After reaction moles of acid = moles of salt

    Therefore pH = pKa = 4.75
    Thanks a lot! I am glad i doubted myself
 
 
 
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