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Pertanyaan

banyaknya CH3COOK (Mr= 98) yang terdapat dalam 100 ml larutan CH3COOK dengan ph = 9,jika diketahui Ka CH3COOH= 10^-5 adalah.....a.0,245 gram b.0,49 gram c.0,98 gr d.9,8 gr e.19,6 gr

2 Jawaban

  • pH = 9
    pOH = 14 - 9
    pOH = 5
    [OH-] = 10^-5

    [OH-] = √Kw / Ka . M
    10^-5 = √10^-14 / 10^-5 . M
    10^-10 = 10^-9 . M
    M = 10^-10 / 10^-9
    M = 10^-1
    M = 0,1 M

    n = M . V
    n = 0,1 . 0,1 L
    n = 0,0,1 mol

    → Banyak yg larut :
    = n . Mr
    = 0,01 . 98
    = 0,98 gram
  • CH3COOH + KOH ---> CH3COOK + H2O
    pH = 9 , pOH = 14-9 =5 , OH- = 10^-5
    OH- = √kw/Ka .[garam]
    10^-5 = √10^-14/ 10^-5 .[garam]
    10^-10 = 10^-9 .[garam]
    [garam] = 10^-10 /10^-9 = 10^-1

    n = M . V = 0.1 × 0.1 = 0.01

    m = n.Mr = 0.01 × 98 = 0,98 gram

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