:mNaCl không đổi=80.15%=12 gam C% dd NaCl sau=12/100.100%=12% mdd sau=200+300=500 gam Tổng mNaCl sau khi trộn=200.20%+300.5%=55 gam C% dd NaCl sau=55/500.100%=11% mdd sau=150 gam mNaOH trg dd 10%=5 gam mNaOH trong dd sau khi trộn=150.7,5%=11,25 gam =>mNaOH trong dd a%=11,25-5=6,25 gam =>C%=a%=6,25/100.100%=6,25% => a=6,25