# Question #d5ccd

Apr 18, 2015

${C}_{1} {V}_{1}$ = ${C}_{2} {V}_{2}$

${C}_{1}$ is original concentration
${V}_{1}$ is original volume
${C}_{2}$ is final concentration and
${V}_{2}$ is the final volume you want to make it up to

So you take the concentration of whatever you want to dilute, say of 1ml volume, then you add 80ml of water.

So then you have a 1:80 dilution.

You can also then work out the concentration of your final dilution by putting in the values into the equation I gave.

Hope I helped :)

Apr 18, 2015

The answer Rogan wrote is correct, but his example is a little off.

Adding 80 mL of water to 1 mL of stock solution will produce a 1:81 dilution, not a 1:80 one.

The ratio between the final volume of the solution, the one you get after diluting the initial sample, and the initial volume of the sample will give you the dilution factor
So,

$\text{DF" = V_"final"/V_"initial}$

If you've got a 1-mL sample, and the dilution factor is 1:80, you'd get

${V}_{\text{final" = "DF" * V_"aliquot" = 80 * 1 = "80 mL}}$

However, this volume includes the volume of the aliquot, the original sample you wanted to dilute, which means that the 80-mL final volume will include 1 mL of sample, the rest being the volume of the diluent

${V}_{\text{diluent" = V_"final" - V_"aliquot" = 80 - 1 = "79 mL}}$

So, to make a 1:80 dilution for a 1-mL sample, you add enough water to get the final volume to 80 mL $\to$ in this example, you would add 79 mL of water to the 1-mL sample.

If you add 80 mL of water, you'll get a dilution of 1:81

$\text{DF} = \frac{80 + 1}{1} = 81$