\(n_{Al_2\left(SO_4\right)_3}=\dfrac{3,42}{342}=0,01\left(mol\right)\)
\(n_{Al\left(OH\right)_3}=\dfrac{0,78}{78}=0,01\left(mol\right)\)
TH1: Kết tủa không bị tan
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3\downarrow+3Na_2SO_4\)
0,005<--------0,03<----------0,01
\(Nxét:0,005< 0,01\Rightarrow Al_2\left(SO_4\right)_3dư\left(t/m\right)\)
\(C_{M\left(NaOH\right)}=\dfrac{0,03}{0,025}=1,2M\)
TH2: Kết tủa tan một phần
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3\downarrow+3Na_2SO_4\)
0,01---------->0,06-------->0,02
\(n_{Al\left(OH\right)_3\left(tan\right)}=0,02-0,01=0,01\left(mol\right)\)
\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)
0,01------->0,01
\(C_{M\left(NaOH\right)}=\dfrac{0,06+0,01}{0,025}=2,8M\)
\(n_{Al_2\left(SO_4\right)_3}=\dfrac{3.42}{342}=0.01\left(mol\right)\)
\(Al_2\left(SO_4\right)_3+6NaOH\rightarrow2Al\left(OH\right)_3+3Na_2SO_4\)
\(n_{Al\left(OH\right)_3}=\dfrac{0.78}{78}=0.01\left(mol\right)\)
\(\Leftrightarrow n_{NaOH}=0.03\left(mol\right)\)
=>Al2(SO4)3 hết, NaOH dư
=>Tính theo Al2SO4 3
\(C_{NAOH}=\dfrac{0.03}{25\cdot10^{-3}}=\dfrac{6}{5}\)