# How do I answer these chemistry hydrate questions?

1. Jun 10, 2007

### Ownaginatious

Since it is getting close to the end of the school year, my teacher has decided to teach really fast :p We started learning about hydrates, and I don't really know if I'm doing the homework right or not because he never gives answers, or takes anything up.

Can someone please walk me through these two questions, and tell me what you get as your answer?

15) What is the volume of water used to make up a 5% by mass solution of sodium acetate when 20 grams of sodium acetate trihydrate is used?

16) How would you prepare 250 mL of 0.100 M barium chloride solution from solid barium chloride dihydrate?

2. Jun 10, 2007

### symbolipoint

15) Calculate the percent sodium acetate in the sodium acetate trihydrate solid; all the rest of your calculations will by weight per weight. This could be like a "percent adjustment" calculation. You start with percent by weight of sodium acetate for the trihydrate, and you want to find how much water (initially by mass, if you want) to add to your compound to obtain your 5% solution.

3. Jun 11, 2007

### Ownaginatious

For the first one I believe I got 229 mL (sorry, can't remember the exact number at the moment) can anyone verify this?

4. Jun 11, 2007

### Kushal

i always get problems with these kind of questions... i'll try it anyway.....

5% of the solution should be NaCH3COO
from 20 g (=0.286 mol) sodiumacetate trihydrate, how much water is present?
1 mol sodiumacetate trihydrate contains 3 mol H2O
0.286 mol = 0.857 mol H2O = 15.4 g H2O
so, mass of sodiumacetate is 20 - 15.4 = 4.57 g

now, 5% should correspond to 4.57 g
95%(water) should correspond to 86.9 g

from the hydrated salt, you already have water... you should be substracting this water from the total water required which then equals to 86.9 - 15.4
= 71.43 g

assuming density of the water to be 1g/cm3.... therefore the volume of water required is 71.43 cm3 = 71.43 mL

now i dunno if my calculations are good... please confirm it

5. Jun 11, 2007

### Kushal

in 1 dm3 there should be 0.1 mol BaCl2
in 250 cm3, there should then be 0.025 mol BaCl2
Mr BaCl2.2H2O is 244
therefore, mass of BaCl2.2H2O required to get 0.025 mol BaCl2 is 6.1 g
from 0.025 mol BaCl2 we will already be getting (0.025*2) = 0.05 mol H2O = 0.9 g, you will need (250 - 0.9) g water more to make up to the 250cm3

so, to prepare 250mL 0.1M BaCl2, you need 6.1g of the hydrated salt + 249.1 cm3 H2O

again, i've assumed density of water to be 1g/cm3

hope i did it right

6. Jun 11, 2007

### chemisttree

Close, but not quite. Show your work and we can help some more.

Kushal's answer for sodium acetate is wrong. His answer for BaCl2 solution is close but not quite the answer.

7. Jun 11, 2007

### Ownaginatious

I thought rather than typing everything out, it would be easier to just show the written work. Sorry if this method of doing the problem is weird, but my teacher wants us to do it in this manner. He really likes ratio tables :p

8. Jun 12, 2007

### chemisttree

I assume that you are using the density of sodium acetate = 1 as is the trihydrate? This is not true but you can make the assumption and not be too far off. (sodium acetate density = 1.528, trihydrate density = 1.45) You are also assuming that there is no volume change due to solution. The volume of sodium acetate trihydrate is 20 mL if you assume that the density is 1.0 g/mL and the volume of the anhydride is 12.06 mL if you assume that the density is 1.0. You have added 20 g (mL) of a solid, 12.06 g (mL) of which is sodium acetate. You have correctly used this value to determine the total volume of solution to yield a 5% concentration. Now all you need to do is subtract out the volumes of sodium acetate (anhydride) and the water of hydration from that total. You only subtracted out the volume of the anhydrous sodium acetate.

Of course, this calculation is off by a small amount since the density of the trihydrate is 1.45 (you assumed 1.0) and you are assuming no change in volume due to solution. Change in volume due to solution is defined here as the difference between the algebraic sum of the volumes of the solvent (l) + solute (s) and the experimentally determined value.

9. Jun 12, 2007

### Ownaginatious

I don't think we're supposed to know the density though...

10. Jun 13, 2007

### chemisttree

How can this possibly be? Somewhere in your arithmetic you subtracted 12.06 grams of anhydrous sodium acetate from 241.12 grams of solution. You then reported your answer in mL! You obviously assumed the density of the solution was equal to water.

But my point was that you ignored the water of hydration when you subtracted the sodium acetate mass from the mass of the solution. You instructed the reader to take 20 grams of sodium acetate trihydrate and add 229.06 mL of water. This would yield 249.06 grams of solution. Your target was 241.12 g, an error of 249.06-241.12=8.04 g The relative error (8/241 * 100 = ~3.3) is what I was trying to address. The untrue assumptions regarding the densities are another source of error.

Like I said, close but not quite the answer.

11. Jun 13, 2007

### Ownaginatious

Oh, I know where the error was. I got 229 as the mass of the water in the solution, but I never took out the mass of the water from the hydrate. The actual answer should be around 221mL I believe...