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    If you put the [OH^-] and Kw values into the equation, you can use the resulting [H^+] to calculate the pH (using pH = -log[H^+].
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    [OH^-] is equal to the concentration of the NaOH, btw.
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    (Original post by crozibear96)
    If you put the [OH^-] and Kw values into the equation, you can use the resulting [H^+] to calculate the pH (using pH = -log[H^+].
    So this would be correct?

    10^-14 = [H^+] [0.1]

    [H^+] = 10^-14 / 0.1

    pH=-log(10^-14/0.1)

    pH= 13


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    (Original post by goodwinning)
    So this would be correct?

    10^-14 = [H^+] [0.1]

    [H^+] = 10^-14 / 0.1

    pH=-log(10^-14/0.1)

    pH= 13


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    The working out looks correct, so it should be right (I don't have a calculator with me to check).
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    Ok, thanks for your help 😄


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    No probies.
 
 
 
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