# Dilution terminology

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?

• It seems your right, but what's OD? Mar 4 '15 at 0:58
• Optical density. Mar 4 '15 at 1:21

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:

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

• so would 50/50 of each be a 2x dilution or.. Mar 4 '15 at 2:07
• Yep, you're right again. Mar 4 '15 at 5:43