n(NaOH)=0.01*0.1=0.001 n(Ba(OH)2) =0.2*0.01=0.002
C(OH-)=(0.001+0.002)/0.01=0.3
所以C(H+)=10^-14/0.3=1/3*10^-13
PH= -lg(C(H+))=13-lg1/3那先算出poh呢？还是不太明白lg是怎么算的POH=-lg( C(OH-))=-lg0.3=1-lg3PH=14-POH =13+ lg3PH=-lg(C(H+))=13-lg1/3 =13+ lg3