Penjelasan:
diketahui:
volume garam=200 ml
konsentrasi garam=0,2 M=2. 10^-1 M
Ka= 2.10^-5
ditanya:
PH=?
jawab:
[OH-] = √kw/ka . mg
= √10^-14 / 2. 10^-5 . 2. 10^-1
= √2. 10^-15 / 2. 10^-5
= √10^-10
= 10^-5
POH = - log [OH-]
= - log 10^-5
= 5
PH = 14-POH
= 14-5
= 9
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Verified answer
Penjelasan:
diketahui:
volume garam=200 ml
konsentrasi garam=0,2 M=2. 10^-1 M
Ka= 2.10^-5
ditanya:
PH=?
jawab:
[OH-] = √kw/ka . mg
= √10^-14 / 2. 10^-5 . 2. 10^-1
= √2. 10^-15 / 2. 10^-5
= √10^-10
= 10^-5
POH = - log [OH-]
= - log 10^-5
= 5
PH = 14-POH
= 14-5
= 9