\[\eqalign{ & \left[ {K3Fe\left( {CN} \right)6} \right]{\text{ }} = \dfrac{{{m_1}}}{{MM \cdot V}} \cr & \left[ {K3Fe\left( {CN} \right)6} \right]{\text{ }} = \dfrac{{1,21}}{{329,2 \cdot 0,775}} \cr & \left[ {K3Fe\left( {CN} \right)6} \right]{\text{ }} = 0,00474{\text{ mol/L}} }\]
b) [K¹+] = ??
Sabemos que são 3 íons K¹+ em cada K3Fe(CN)6:
\[\eqalign{ & \left[ {K + } \right]{\text{ }} = {\text{ }}3{\text{ }} \cdot {\text{ }}\left[ {K3Fe\left( {CN} \right)6} \right] \cr & \left[ {K + } \right]{\text{ }} = {\text{ }}3 \cdot {\text{ }}4,74{\text{ }} \cdot {\text{ }}{10^{ - 3}} \cr & \left[ {K + } \right]{\text{ }} = {\text{ }}1,89{\text{ }} \cdot {\text{ }}{10^{ - 2}}{\text{ mol/L}} }\]
c)
\[\eqalign{ & \left[ {Fe\left( {CN} \right)6 - } \right]{\text{ }} = {\text{ }}\left[ {K3Fe\left( {CN} \right)6} \right] \cr & \left[ {Fe\left( {CN} \right)6 - } \right]{\text{ }} = {\text{ }}4,74{\text{ }} \cdot {\text{ }}{10^{ - 3}}{\text{ mol/L}} }\]
d)
\[p = 0,156\%\]
e)
\[mmol=2,37 \cdot {10^{ - 4}}{\text{ mol}}\]
f)
\[1560 ppm\]
g)
\[pk=7,8\]
h)
\[pFe(CN)6=0,18\]
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