Strong acid and strong base calculation hwk

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    Hi,

    There are a couple of questions about acids and bases that I don't understand.

    1. Calculate the pH of water at 50 degrees when Kw= 5.48 x 10^-14. Is the water neutral? Explain your answer.

    2. 3.5g of impure NaOH (98.7%) dissolved in water and made up to 100cm^3. Then 25cm^3 of 0.35moldm-3 diprotic acid is added. Calculate the pH. I've looked at a few explanations for this question but I'm still confused.

    Thanks
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    (Original post by VioletPhillippo)
    Hi,

    There are a couple of questions about acids and bases that I don't understand.

    1. Calculate the pH of water at 50 degrees when Kw= 5.48 x 10^-14. Is the water neutral? Explain your answer.

    2. 3.5g of impure NaOH (98.7%) dissolved in water and made up to 100cm^3. Then 25cm^3 of 0.35moldm-3 diprotic acid is added. Calculate the pH. I've looked at a few explanations for this question but I'm still confused.

    Thanks
    1. Do you know the definition of pH? Do you know the definition of kw?

    2. Work out what is left behind after the neutralisation reaction.
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    I did this in lesson today... I have used the method my teacher gave me but I'm not sure whether the pH is given as 7 or if you have to find that?
    Hopefully this is somewhat useful anyway

    1. Calculate the pH of water at 50 degrees when Kw= 5.48 x 10^-14. Is the water neutral? Explain your answer.

    I am assuming the water pH would be 7. So you do 10 to the power of -7 which gives you 1x10^-7

    you then rearrage the Kw equation (Kw= [H+] [OH-]) to OH- = 5.48x10^-14 / 1x10^-7

    This equals, 5.48x 10^-7 which would make the water slightly acidic.
 
 
 
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