# conc. calculation - (Mar/05/2007 )

Greetings

not a homework question (part of lab experiment)-just not sure where to but it and I don't know how to work it out...

we have 2ml of distilled water + 8ml 0.154M NaCl...

how do I work out the new conc?....

also how do i make up 0.075M NaCl if I have distilled water available and 0.154M NaCl available--apparently our group came up with 5ml of each-but I don't know how to work that out either...

cheers
biology_06er

-biology_06er-

Your best bet could be to use c1.v1=c2.v2 . For Q1, you have c1 (conc 1), v1 (vol 1) and v2. For Q2, you know what your c1, v1 and c2 are. Re-arrange the equations to solve for the other variable and that will give you the final concentration or volume, as required.

-swanny-

the new concentration of NaCl.

an other way to calculate it, if you prefer, is to think about how much the NaCl has been diluted.
the volume was first 8 mL, and now it's diluted to 2+8 = 10 mL

so you diluted your NaCl 10/8 times.
the new concentration is 0.154 / (10/8) = 0.154 *8 / 10 = 0.1232M

to prepare 0.075M :
As you can see, 0.075 M is quite the half of 0.154M, so if you dilute twice (5 mL of NaCl and 5 mL of water), you will have quite 0.075M

if you want to be more precise, you can do as Swanny suggested.
Let's assume you need 10 mL, so V2 is 10 mL C2 is 0.075M and C1 is 0.154M
you want to know the initial volume of NaCl to dilute in 10 mL : V1

V1 = (C2*V2)/C1 = 0.075*10/.154 = 4.87 mL

so you put 4.87 mL of NaCl and 5.13 mL of water.

-Missele-