If I mix a $\pu{2 M}$ solution with a different $\pu{1 M}$ solution, do the individual solutions reduce in molarity to $\pu{1.33 M}$ and $\ce{0.33 M}$ respectively?

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    $\begingroup$ Depends on the mixing ratio. Are you mixing them 50:50 or in a different ratio? $\endgroup$ – Philipp Mar 2 '14 at 21:12
  • $\begingroup$ @Philipp, good question. I am unsure. The machine dispenses them for me. I will check and edit the question accordingly. $\endgroup$ – user2679447 Mar 2 '14 at 21:14
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    $\begingroup$ This will all depend on what you are mixing. If you are mixing 2 M NaOH with 1 M NaOH, you will have a new concentration somewhere between 1 and 2 M depending on the volumes. If you mix 2 M NaOH with 1 M NaCl, you have a different scenario. If you are mixing 2 M NaOH with 1 M HCl, then you have a reaction, which will be a third type of scenario. $\endgroup$ – Ben Norris May 2 '14 at 1:38

You need to know how much of each solution you add and the final total volume. The molarity of each one would be

$$\frac{C_1 \cdot V_1}{V_1 + V_2} \quad \text{and} \quad \frac{C_2 \cdot V_2}{V_1 + V_2}$$

where $C_1 = \pu{2 M}$, $C_2 = \pu{1 M}$, $V_1$, $V_2$ - volumes of $\pu{2 M}$ and $\pu{1 M}$ solutions, respectively.


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