0.1mol/L的氯化铵溶液PH是多少?
问题描述:
0.1mol/L的氯化铵溶液PH是多少?
答
NH4++H20=NH3H2O+H+
[NH3H2O][H+]/[NH4+]=1.8×10^-5
[H+]^2=1.8×10^-5*0.1
PH=-LOG(SQRT(1.8*10^(-5)*0.1))=2.87