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Pertanyaan

diketahui 100 ml larutan NaOH 0,002 mol , pH larutan tersebut adalah...

A.12
B.10
C.9
D.7
E.4

2 Jawaban

  • M = n/V
    = 0,002 mol/0,1 liter
    = 2.10^ -2 M

    NaOH → Na^+ + OH^-
    2.10^-2 M ~~~~~~ 2.10^-2 M

    pOH = -log[OH^-]
    = -log[2.10^-2]
    = 2-log 2
    pH + pOH = 14
    pH = 14 - ( 2-log2 )
    = 12 + log 2
  • M NaOH =
    Mol / V = 0.002mol /0,1 liter = 0.02 M
    (OH-) = M NaOH = 0.02
    pOH = - log (OH-) = - log 0.02 = 2 - log 2
    pH = 14 - pOH = 14 - 2 + log 2
    = 12 + log 2

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