Kimia

Pertanyaan

hitung ph larutan nh4oh 0,001 m ( kb = 10-5 )

1 Jawaban

  • [OH-] = √kb . [NH4OH]
    [OH-] = √10^-5 . (1 x 10^-3)
    [OH-] = √10^-5 . 10^-3
    [OH-] = √10^-8
    [OH-] = 10^-4

    pOH = - log [OH-]
    pOH = - log [10^-4]
    pOH = 4

    pH = 14 - pOH
    pH = 14 - 4
    pH = 10

Pertanyaan Lainnya