Chemistry. dilution?

Question

Calculate the new molarity if each of the following dilutions is made. assume the volumes are additive.

a) 25.0 mL of water is added to 10.0 mL of 0.251 M CaCl2 solution

b) 25.0 mL of 1.25 M NaCl solution is diluted with an equal volume of water

please explain! thanks

Answer 1

Or you can use the unit-factor method and a conversion factor.  Since the molarity is a ratio, it works just like a conversion factor.

Start with the original volume, multiply by the molarity to get the moles of solute, then divide by the new volume to get the new molarity.  Easy.

10.0 mL x (0.251 mol CaCl2 / 1L) x (1 / 35.0 mL) = 0.0717 M CaCl2

The second one is trivial.  The concentration is cut in half.  You get 0.625 M NaCl

Or if you would like to see it again...

25.0 mL x (1.25 mol NaCl / 1L) x (1 / 50.0 mL) = 0.625 M NaCl

Answer 2

You can use:

C1V1 = C2V2

Where

C1 = concentration of original solution

V1 = volume of the original solution used

C2 = concentration of final solution

V2 = volume of final solution

You must keep all the units constant for this equation to work. So if V1 is in ml then V2 must also be in ml.

a)

C1 = 10.0 ml

V1 = 0.251 M

V2 = 35.0 ml (25.0 ml water + 10.0 ml CaCl2 solution)

C2 = unknown

So C2 = V1C1 / V2

= (0.251 (M) x 10.0 ml) / 35.0 ml

= 0.0717M

b) you have doubled the volume therefore halved the conc.

= 0.625 M

you don't really need the equation for this one, but here it is for the sake of learning to use it.

C1 = 1.25 M

V1 = 25.0 ml

C2 = unknown

V2 = 50.0 ml

C2 = C1V1 / V2

= (1.25 M x 25.0 ml) / 50.0 ml

= 0.625 M