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For thing like 8× dilution, that would be like 1 ml of a solution and then 7 ml of something like water to dilute right? Would 50 ml solution and 50 ml water be a 2× dilution? I’m a bit unsure about the terminology. If I an OD of 0.123 from a sample diluted 5×, would multiply 0.123 by 5 to get the real OD?

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  • $\begingroup$ It seems your right, but what's OD? $\endgroup$
    – Mithoron
    Mar 4 '15 at 0:58
  • $\begingroup$ Optical density. $\endgroup$
    – Mar09
    Mar 4 '15 at 1:21
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You are right. The × in 8× refers to the fold-change in concentration between the concentrated source and diluted solution.

$$\mathrm{DF = \frac{C_{stock}}{C_{working}}}$$ where $\mathrm{C_{stock}}$ and $\mathrm{C_{working}}$ are the concentrations of a stock solution and the desired final concentration, respectively.

The concentration of the working solution can be expanded:

$$\mathrm{C_{working} = \frac{C_{stock} V_{stock}}{V_{working}} = \frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}}$$ where $\mathrm{V_{stock}}$ is the volume of the stock solution and $\mathrm{V_{diluent}}$ is the volume of diluent (e.g. pure water) added.

Plugging back into the DF equation:

$$\mathrm{DF = \frac{C_{stock}}{\frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}} = \frac{V_{stock}+V_{diluent}}{V_{stock}}}$$

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    $\begingroup$ so would 50/50 of each be a 2x dilution or.. $\endgroup$
    – Mar09
    Mar 4 '15 at 2:07
  • $\begingroup$ Yep, you're right again. $\endgroup$
    – Curt F.
    Mar 4 '15 at 5:43

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