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How can I calculate the final concentration of two ions in a river after the confluence of two rivers?

  • River 1 flows at $\small\sf{7000\ L/s}$, contains $\small\sf{500\ mg/L\ Mg,\ 1.1\ g/L\ Ca}$
  • River 2 flows at $\small\sf{11\ m^3/s}$, contains $\small\sf{0.34\ g/L\ Mg,\ 200\ mg/L\ Ca}$

How do I find Mg and Ca concentrations in River 3, after the joining of Rivers 1 and 2?

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You have to use mass weighted averaging to calculate the final concentrations of elements. The equation for your situation would be:

$\sf\bar{q}\ =\ \dfrac{m_1.q_1\ +\ m_2.q_2}{m_1\ +\ m_2}\ =\ \dfrac{\rho_1.v_1.q_1\ +\ \rho_2.v_2.q_2}{\rho_1.v_1\ +\ \rho_2.v_2}$

where:

  • $\small\sf m$ is the mass flow rate
  • $\small\sf v$ is the volume flow rate
  • $\small\rho$ is the density of water
  • $\small\sf q$ is the quantity/concentration of the element

Every second, $\small\sf{7000\ L}$ of water flows from River 1 into River 3. Similarly, $\small\sf{11\ m^3}$ of water flows from River 2 into River 3. The combined inflow into River 3 is $\small\sf{18\ m^3}$ every second.

Assuming the density of water in both inflowing rivers is the same, volume of water can be used instead of mass; if not then the density of water for each river needs to be used.

Assuming the water density of both rivers is the same, convert all to units SI units and the average concentration of Mg in River 3 is,

$\sf{\dfrac{7(0.5) + 11(0.34)}{(7 + 11)}}$ = $\sf{0.402\ g/L\ Mg}$

I'll let you do the calculation for Ca.

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