\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
0,2------->0,1--------->0,1
Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư
Trong dung dịch D có:
\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)
b
\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)
0,35<---------0,35
\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)