[MissDebbie]

How much water must be added to 125 mL of a 30% NaOH solution in order to prepare a 7% solution?

I multiplied 125 mL by 30 and divided the answer by 7 and got an answer of 535.7.  I then subtracted 125.0 mL of water from the 535.7 mL.  Is this correct?  I am not sure whether I was supposed to subtract the 125.0 at the end?

[Mitch]

I assume you used M1V1=M2V2?

[MissDebbie]

V₂=V₁ x C₁/C₂ is the formula I used.

[Borek]

V₂=V₁ x C₁/C₂ is the formula I used.

Multiply both sides by C₂

http://www.chembuddy.com/?left=concentration-questions&right=molarity-dilution-q1

and

http://www.chembuddy.com/?left=concentration-questions&right=molarity-dilution-q2

Your answer is (almost) correct - to be precise you need 547.2mL (taking density changes into account).

Try CASC for such calculations.

[MissDebbie]

Thank you for guiding me on this.  I have a test on these concepts on Monday morning at 8 AM.   I am still a little confused with your last comment.  In order to complete this problem, would we have to subtract 125 mL from the final total or is it left as is.  I keep trying to figure this out in my head and I'm having difficulty.

[Borek]

Oops, sorry - somehow I managed to calculate correct result for percentage, but to not spot that equation you used is valid only for molar concentrations

To calculate properly amount of water you need to add you have to know density of the 30% and 7% solutions. Why? Because mass percentage is defined as weight of solute/weight of solution (see equation 1.1 in my concentration lectures).

Densities are usually given in tables (for NaOH you will find density table on my website too, in the CASC description). In your case - 30% solution has a density of 1.33 g/mL and 7% - 1.08 g/mL.

Your original solution has a mass of 125*1.33 = 166.3 g. In this solution there is 30/100*166.3 = 49.88 g NaOH. It will be 7% (or 0.07) of the mass of the final solution - so the final solution mass is 49.88/0.07 = 712.5 g.

Amount of water that have to be added is the difference between the final mass and the starting mass - 712.5-166.3 = 546.2 g. To be very precise you should take water density into account, but if you assume that 1 mL of water is just 1 g you will not make large error - so you have to add 546.2 mL of water (my previous answer was a little bit more precise, as it took into account more significant digits and the real water density).

Your original approach will not give correct result in case of mass percentages and concentrated solutions!