Today is 11 March 2010 -- We are studying solution chemistry

You can study this information on your own:  It is Chapter 13 in your text (HOLT, Chemistry).  Read and take notes in your lab notebook on Ch. 13, sections 1 and 2.  To prepare for this section, define the key terms on page 454 of your text.  Also, answer the following questions:

1.  We need a 30% solution of copper sulfate.  We have available 40 ml of 25% copper sulfate and an unlimited amount of 60% CuSO₄.  How many milliliters of the 60% solution should we mix to get the required 30% concentration?

.3(40ml + x) = .25(40ml) + .6x

12ml + .3x = 10ml + .6x

12ml - 10ml = .6x - .3x

2ml = .3x

2ml/.3 = x = 6.67 ml

How do I know this?

A "30%" solution means 30 grams of a solute in 100 ml of a solvent.  Similarly, "25%" is 25 grams in 100 ml.  Since we have 40 ml, then the amount of solute we have is:

 (25 grams) x (40 ml/100ml) = 10 grams

The 60% solution is 60 grams in 100 ml, but we don't know how much to add so we write:

(60 grams) x (x ml/100 ml) = 0.6x g

The result, 30%, means 30 grams in 100 ml. But we only know the volume as (40 ml + x ml) so we write: (30 grams /100 ml) x (40ml + x ml)

Put it together:

10 g + (.6x g) = (.3g/ml)x(40ml + x ml)

10 g + .6x g = 12 g + .3x g

Rearrange:

.6x g - .3x g = 12 g - 10 g

.3x g = 2 g

x = 2 g/.3g = 6.67

Prove this!

The atomic mass of CuSO₄ = 63.57 amu + 32.06 amu + (4)(15.99 amu) = 159.59 amu.

The number of moles of each solution is:

Conc	Moles/ 100 ml	Moles/ml	ml need  (or have)	moles
30%	30/159.59	0.00188	40 + x	0.0752+ 0.00188x
25%	25/159.59	0.00157	40	0.0628
60%	60/159.59	0.00376	x	0.00376x

0.0752 + 0.00188x = 0.0628 + 0.00376x

Rearrange

0.0752 - 0.0628 = 0.00376x - 0.00188x

0.0124 = 0.00188x

x= 0.0124/0.00188 = 6.60 ml of 60% solution

2.  It turns out that we need 140 ml of the 30% copper sulfate solution.  Did we make enough in the first problem?  If not, how many milliliters of each solution (25% and 60%) do we need to make 140 ml of 30% CuSO₄ solution?

Let x be the ml needed of one solution.  Then 140 - x = ml of other solution.

(140ml)(30%) = (25%)(x) + (60%)(140ml - x)

4200ml% = 25x% + 8400ml% - 60x%

4200ml% - 8400ml% = 25x% - 60x%

-4200ml% = -35x%

x = 4200ml%/35% = 120 ml (of 25% solution)

140 - x = 40 ml of 60% solution