Jawabanmu

2014-02-18T08:01:49+07:00
P(OH-)= -log 2 x 10^-3
P(OH-)= 3-log2 =2,699
P(H+)= 14-2,699=11,301
2014-02-18T16:42:27+07:00
[OH] = 2 x 10⁻³  pOH = - log 2 x 10⁻³  = 3 - log 2 = 3 -0,301 = 2,699
pH = 14 -2,699  = 11,301