jika pH = 5+2log2, berapa [H+] ?

ph = 5+2log2 = 5-log2^-2 = 5-log1/2^2 = -log(1/4)^-5
ph = -log[H+] ,
maka [H+] = (1/4)^-5

2log2=1
ph nya 5+1=6
Karna ph = -log [H]
Jaadi [H]= 10 ^ -6