Question

Ph larutan yang mengandung 6 gram CH3COOH (Mr=60) dan 0,1 mol CH3COONA (Ka=1,0 x 10^-5) adalah....

Answer 1

N= 6/60= 0,1 konsentrasi H = ka x 0,1/0,1 [H] = 1,0 x 10 pangkat -5 x1 [H]= 1,0 x 10 pangkat-5 ph= -log 1,0 x 10 pangkat -5 ph=5-log 1,0 ph=5+0 ph= 5 kalau ga salah begitu hehe

Answer 2

N(mol) CH3COOH = gr/mr = 6/60 = 0,1 mol -->na
n CH3COONa = 0,1 mol -->ng
[H]=ka x na/ng
    =1.10^-5 x 0,1/0,1
    =1x 10^-5
pH=-log [H]
    =- log 1 x 10^-5
    =5