Hitunglah ph larutan CH3COONa 1 M (ka ch3cooh = 1x10 pangkat -5;kb nh4oh=1x10-5

ini soal hidrolisis garam ,,, kemungkinan soalnya CH3COONa 0,1 M .. bukan 1M
karna ini garam sifat basa ,, maka
[OH-] = AKAR [ (KW / Ka) x ( konsentrasi CH3COONa ) ]
[OH-] = AKAR [ (10^-14 / 10^-5 ) x ( 0,1M ) ]
[OH-] = AKAR [ 10^-10 ]
[OH-] = 10^-5
pOH = 5 ,, Maka PH = 9

kalau CH3COONa 1 M maka
[OH-] = AKAR [ 10^-9 ]  ,,, {OH }= 10^-4,5 .. pOH = 4,5 ,, pH = 9,5