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Pertanyaan

Larutan asam asetat (Ka = 2*10^-5) yang mempunyai pH yang sama dengan larutan 2*10^-3 molar asam klorida, mempunyai konsentrasi...

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  • pH CH3COOH = pH HCl

    HCl : Asam kuat
    [H+] = M x val
    = 2 . 10^-3 x 1
    = 2 . 10^-3 M

    pH HCl = - log [H+]
    = - log 2 . 10^-3
    pH HCl = 3 - log 2 = pH CH3COOH

    pH CH3COOH = - log [H+]
    3 - log 2 = - log [H+]
    [H+] = antilog 3 - log 2
    = 2 . 10^-3 M

    CH3COOH : Asam lemah
    [H+] = akar (Ka x M)
    2 . 10^-3 = akar (2 . 10^-5 x M)
    (2 . 10^-3)^2 = 2 . 10^-5 x M
    4 . 10^-6 = 2 . 10^-5 x M
    M = (4 . 10^-6)/ (2 . 10^-5)
    M CH3COOH = 2 . 10^-1 M

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