pH dari 100ml larutan CH3COOH 0,009 M adalah.....(Ka CH3COOH=10^-5 A.4-log3 b.6+log3 c.8-log3 d.8+log3 e.12-log9

V ( Ch3COOH ) = 100 ml
a ( CH3COOH) = 9 x 10^-3
Ka = 10^-5
Dit : PH ?
PH = - log ( √Kaxa )
= -log ( √10^-5x9.10^-3 )
= - log ( √9x10^-8 )
= - log 3x10^-4
= 4 - log 3 ( A )
semoga terbantu

Jadi pH nya adalah A. 4-log 3