nNaOH (dd1M)=0,3mol; nNaOH (dd1,5M)=0,2.1,5=0,3mol;
nNaOH (dd sau)=nNaOH (dd1M) + nNaOH (dd1,5M)=0,3+0,3=0,6mol
Vdd sau= 300+200=500ml=0,5 lít
CM(ddsau)=0,6/0,5=1,2M
m dd sau=d.V(ddsau)=500.1,05=525g;
mNaOH=0,6 .40=24g
C% ddsau=24/525 .100% ~=4,57%
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