\(n_{Al\left(OH\right)_3}=\dfrac{3,9}{78}=0,05mol\)
\(n_{NaOH}=0,2mol\)
\(n_{KOH}=0,2.1,5=0,3mol\)
\(\rightarrow n_{OH^-}=0,2+0,3=0,5mol\)
\(n_{H^+}=n_{HCl}=0,05mol\)
\(n_{Al^{3+}}=0,1xmol\)
H++OH-\(\rightarrow\)H2O
\(n_{OH^-\left(pu\right)}=n_{H^+}=0,05mol\)
\(n_{OH^-\left(dư\right)}=0,5-0,05=0,45mol\)
Al3++3OH-\(\rightarrow\)Al(OH)3(1)
Al(OH3+OH-\(\rightarrow\)AlO2-+H2O(2)
- Sau phản ứng (2) Al(OH)3 còn dư 0,05 mol
\(n_{OH^-\left(1\right)}=3n_{Al^{3+}}=0,3xmol\)
\(n_{Al\left(OH\right)_3\left(1\right)}=n_{Al^{3+}}=0,1xmol\)
\(n_{Al\left(OH\right)_3\left(2\right)}=0,1x-0,05\) mol
\(n_{OH^-\left(2\right)}=n_{Al\left(OH\right)_3\left(2\right)}=0,1x-0,05\)
Theo PTHH (1,2) ta có:
\(n_{OH^-}=n_{OH^-\left(1\right)}+n_{OH^-\left(2\right)}\)
\(\rightarrow\)0,45=0,3x+0,1x-0,05\(\rightarrow\)0,4x=0,5\(\rightarrow\)x=1,25M
