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20 February, 16:07

The Kw of pure water at 40°C is 2.92 x 10^-14.

(a) Calculate the [H+] and [OH - ] in pure water at 40°C.

(b) What is the pH of pure water at 40°C?

(c) If the hydroxide ion concentration in a solution is 0.10 M, what is the pH at 40°C?

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  1. 20 February, 18:18
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    a) [H+]=[OH-] = 1.708e-7

    b) pH = 6.76

    c) pH=13

    Explanation:

    a) Kw=[H+] (OH-) = 2.92e-14

    As [H+]=[OH-]=take it equal to x

    x2=2.92e-14

    take squre root on both sides

    we get x=1.708e-7=[H+]=[OH-]

    b) As we know that pH = - log[H+]

    putting the value in above equaton we get

    pH = - log[1.708e-7]

    pH = 6.76

    c) if [OH-]=0.10 M FIND pH

    pOH = - log (OH-)

    pOH = - log (1e-1)

    pOH = 1 but we also know that pH+pOH=14

    sp pH=14-pOH

    pH=14-1

    pH=13
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