\(n_{NaOH}\left(dd1\right)=\dfrac{20}{40}=0,5\left(mol\right)\)
\(n_{NaOH}\left(dd2\right)=0,25.1,5=0,375\left(mol\right)\)
\(Vddsau=0,8+1,5=2,3(l)\)
\(n_{NaOH}\left(sau\right)=0,5+0,375=0,875\left(mol\right)\)
CM của dung dịch sau khi trộn .
\(C_{M_{NaOH}}\left(sau\right)=\dfrac{0,875}{2,3}\approx0,38\left(M\right)\)