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i've been sitting here trying to figure this problem out and I have no idea what to do..

A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock solution is added to 50.00 mL of water. Calcuate the concentration of ammonium ions and sulfate ions in the final solution.

By Cherrybarry (Cherrybarry) on Sunday, October 10, 2004 - 10:04 am: Edit

I don't have my periodic table with me so I will just have to explain.

Find out how many moles or milimoles of the solute you have. This is how much is dissolved in 100 ml. So divide by 10 to find out how many moles or milimoles in 10 ml.

The [] of the solute is this number (in milimoles) divided by 60, since you have 60 ml of water. Or you can use .06 if you are using moles. Then convert into ions. 1 M solute = 2 M amm. ions and 1 M sol. = 1 M sulfate ions.

By Adingbobo (Adingbobo) on Sunday, October 10, 2004 - 10:52 pm: Edit

hmm.. i think i followed your directions but i didn't get the right answer. here's what i did:

10.8 g ammonium sulfate x 1 mol/228.42 g =
= 0.0473 mol ammonium sulfate

then i divided 0.0473 mol by 10 to get
0.00473 mol

0.00473 mol / 60 mL x 1000 mL/ 1 L =
= .0788 M

and I divided .0788 by 2 and I got .0394 M for ammonium

my answer key says .272 for ammonium. what did i do wrong?

By Adingbobo (Adingbobo) on Sunday, October 10, 2004 - 11:43 pm: Edit

nevermind.. my molar mass is wrong =*( thanks to athlonmj for pointing that out.. hehe