Tentukan pH larutan asam lemah CH3COOH 0,1M (Ka CH3COOH = 1 × 10-5)

diket: Ka= 10^-5, Ma = 10^-1
ditanya: [H+]
jawab



[tex] \sqrt{ka \times ma} \\ \sqrt{ {10}^{ - 5} \times {10}^{ - 1} } \\ \sqrt{ {10}^{ - 6} } \\ {10}^{ - 3} [/tex]

jawabannya adalah.....

Ka CH3COOH=1,8×10^-5.


[H+] =√ ka.Ma
=√1,8× 10^-5 . 1× 10^-¹
=√1.8× 10^-6
=1,34× 10^-³

PH= -Log [H+]
=- Log 1,34× 10^³
=3 - Log 1,34
=3 - 0,12
= 2,88.

jadi,PH larutan CH3COOH adalah 2,88.

semoga bermanfaat bagi anda...