Respostas
\[{K_f}{\alpha _4} = \left( {4,9 \times {{10}^8}} \right) \times 0,36 = 1,764 \times {10^8} = \dfrac{{\left[ {{\text{Mg}}{{\text{Y}}^{2 + }}} \right]}}{{\left[ {{\text{M}}{{\text{g}}^{2 + }}} \right]{C_{EDTA}}}}\]
Início:
\[\eqalign{ & {\text{n^\circ mols }}{{\text{Y}}^{{\text{4 - }}}} = \left( {0,250 {\text{mol/L}}} \right) 5 {\text{mL}} = 1,25 {\text{mmol}} \cr & }\]
\[{\text{n^\circ mols M}}{{\text{g}}^{{\text{2 - }}}} = 0,0250 {\text{mol/L}} \times 20 {\text{mL = 0,5}} {\text{mmol}}\]
Reagem:
\[{{\text{C}}_{EDTA}} = \dfrac{{1,25{\text{mmol}}}}{{25{\text{mL}}}} = 0,05 {\text{mol/L}}\]
\[{{\text{C}}_{EDTA}} = \dfrac{{1,25{\text{mmol}}}}{{25{\text{mL}}}} = 0,05 {\text{mol/L}}\]
\[\left[ {{\text{Mg}}{{\text{Y}}^{2 + }}} \right] = \dfrac{{0,5 mmol}}{{25 mL}} = 0,02 {\text{mol/L}}\]
Donde: \(\left[ {{\text{M}}{{\text{g}}^{2 + }}} \right] = 4,41 \times {10^{ - 8 }}{\text{mol/L}}\)
\[{\text{pMg = - log}}\left[ {{\text{M}}{{\text{g}}^{2 + }}} \right] = 7,35\]
Responda
Para escrever sua resposta aqui, entre ou crie uma conta