• Title/Summary/Keyword: A330

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The Chemical Composition of the Nagdong River Downstream Water (낙동강 하류수의 수질조성에 대하여)

  • WON Jong Hun;LEE Bae Jung
    • Korean Journal of Fisheries and Aquatic Sciences
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    • v.14 no.2
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    • pp.47-58
    • /
    • 1981
  • Relationships between the electrical conductivity and the contents of the chloride, sulfate, calcium, magnesium, sodium, potassium and total major inorganic ions, and between each, chemical conservative constituents were calculated with the data which sampled at the lesions of Mulgeum and between Namji and Wondong from March 1974 to April 1980. Semilogarithmic relations were found between the electrical conductivity and the contents of monovalent ions, and logarithmic relations were found between the electrical conductivity and the contents of divalent ions at the both regions. The relational equations between the electrical conductivity $\lambda_{25}$and the contents of the major inorganic ions at Mulgeum are as follows: $log\;Cl(ppm)\;=\;2.37{\cdot}\lambda_{25}(m{\mho}/cm)+0.733{\pm}0.141$, $log\;SO_4(ppm)=1.12{\cdot}log\lambda_{25}(m{\mho}/cm)+2.14{\pm}0.18$, $log\;Ca(ppm)=0.615{\cdot}log\lambda_{25}(m{\mho}/cm)+1.67{\pm}0.12$, $log\;Mg(ppm)=0.756{\cdot}log\lambda_{25}(m{\mho}/cm)+1.27{\pm}0.11$, $log\;Na(ppm)=2.82{\cdot}\lambda_{25}(m{\mho}/cm)+0.551{\pm}0.133$, $log\;K(ppm)=1.33{\cdot}\lambda_{25}(m{\mho}/cm)+0.136{\pm}0.095$, and total inorganic ions $C(ppm)=399{\cdot}\lambda_{25}(m{\mho}/cm)-0.9{\pm}14.6$. The relational equations between the electrical conductivity ($\lambda_{25}$) and the contents of the major inorganic ions at the region between Namji and Wondong a.e as follows: $log\;Cl(ppm)=4.27{\cdot}\lambda_{25}(m{\mho}/cm)+0.380{\pm}0.138$, $log\;SO_4(ppm)=0.915{\cdot}log\lambda_{25}(m{\mho}/cm)+1.95{\pm}0.18$, $log\;Ca(ppm)=0.756{\cdot}log\lambda_{25}(m{\mho}/cm)+1.74{\pm}0.12$, $log\;Mg(ppm)=1.00{\cdot}log\lambda_{25}(m{\mho}/cm)+1.41{\pm}0.10$. $log\;Na(ppm)=2.47{\cdot}\lambda_{25}(m{\mho}/cm)+0.614{\pm}0.065$, $log\;K(ppm)=1.62{\cdot}\lambda_{25}(m{\mho}/cm)+0.030{\pm}0.060$, and total inorganic ions $C(ppm)=323{\cdot}\lambda_{25}(m{\mho}/cm)+11.7{\pm}9.3$. Logarithmic relations were found between each chemical conservative constituents at Mulgeum and the equations are as follows: $log\;Cl(ppm)=0.711{\cdot}log\;SO_4(ppm)+0.488{\pm}0.206$, $log\;Cl(ppm)=0.337{\cdot}log\;Ca(ppm)+0.822{\pm}0.130$, $log\;Cl(ppm)=0.605{\cdot}log\;Mg(ppm)-0.017{\pm}0.154$, $Cl(ppm)=0.676{\cdot}Na(ppm)+2.31{\pm}4.67$, $log\;Cl(ppm)=0.406{\cdot}log\;K(ppm)-0.092{\pm}0.112$, $log\;SO_4(ppm)=0.378{\cdot}log\;Ca(ppm)+0.721{\pm}0.125$, $log\;SO_4(ppm)=0.462{\cdot}log\;Mg(ppm)+0.107{\pm}0.118$, $log\;SO_4(ppm)=0.592{\cdot}log\;Na(ppm)+0.313{\pm}0.191$, $log\;SO_4(ppm)=0.308{\cdot}log\;K(ppm)-0.019{\pm}0.120$, $Ca(ppm)=0.262{\cdot}Mg(ppm)+0.74{\pm}1.71$. $log\;Ca(ppm)=1.10{\cdot}log\;Na(ppm)-0.243{\pm}0.239$, $Ca(ppm)=0.0737{\cdot}K(ppm)+1.26{\pm}0.73$, $log\;Mg(ppm)=0.0950{\cdot}Na(ppm)+0.587{\pm}0.159$, $log\;Mg(ppm)=0.0518{\cdot}K(ppm)+0.111{\pm}0.102$, and $Na(ppm)=0.0771{\cdot}K(ppm)+1.49{\pm}0.59$. Logarithmic relations were found between each chemical conservative constituents except a relationship between the chloride and calcium contents at the region between Namji and Wondong, and the equations are as follows : $log\;Cl(ppm)=0.312{\cdot}log\;SO_4(ppm)+0.907{\pm}0.210$, $log\;Cl(ppm)=0.458{\cdot}log\;Mg(ppm)+0.135{\pm}0.130$, $Cl(ppm)=0.484{\cdot}logNa(ppm)+0.507{\pm}0.081$, $Cl(ppm)=0.0476{\cdot}K(ppm)+1.41{\pm}0.34$, $log\;SO_4(ppm)=0.886{\cdot}log\;Ca(ppm)+0.046{\pm}0.050$, $log\;SO_4(ppm)=0.422{\cdot}log\;Mg(ppm)+0.139{\pm}0.161$, $log\;SO_4(ppm)=0.374{\cdot}log\;Na(ppm)+0.603{\pm}0.140$, $log\;SO_4(ppm)=0.245{\cdot}log\;K(ppm)+0.023{\pm}0.102$, $log\;Ca(ppm)=0.587{\cdot}log\;Mg(ppm)+0.003{\pm}0.088$, $log\;Ca(ppm)=0.892{\cdot}log\;Na(ppm)+0.028{\pm}0.109$, $log\;Ca(ppm)=0.294{\cdot}log\;K(ppm)-0.001{\pm}0.085$, $log\;Mg(ppm)=0.600{\cdot}log\;Na(ppm)+0.674{\pm}0.120$, $log\;Mg(ppm)=0.440{\cdot}log\;K(ppm)+0.038{\pm}0.081$, and $log\;Na(ppm)=0.522{\cdot}log\;K(ppm)-0.260{\pm}0.072$.

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