Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Calculate the concentration of ammonium ions and sulfate ions in the final solution.?
A solution is prepared by dissolving 11.0 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 12.50 mL sample of this stock solution is added to 54.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
NH4+
_________M
SO42-
________M
2 Answers
- Isabel DLv 41 decade agoFavorite Answer
(NH4)2SO4 --> NH4+ + SO42-
m.w. = 132.07 g/mol
11.0g / 132.07 = 0.083 moles / 0.1L = 0.83M original solution
0.0125L x 0.83M = 0.0104 moles removed from stock solution. 0.0104moles / 0.054L water = 0.192M new concentration of ammonium sulfate.
Since there are 2 moles of ammonium ions in each mole of the molecule, the new concentration of ammonium ions will be 0.384M, and 0.192M for sulfate ions.
- blancLv 44 years ago
Step a million: convert grams of (NH4)2 SO4 into moles (hint: use the molecular weight) Step 2: comprehend that as quickly because it dissolves, it is going to sort 2 NH4+ ions and purely one SO4 -- ion Step 3: calculate the volum (in LITERs) formed via the answer (hint: 12 + 51; then convert from mL to L) Step 4: Calculate the concentration of SO4 -- via dividing the sort of mole via the sort of Liters (this ought to be glaring because of the fact the gadgets for concentration are mole / L Step 5: are you able to discover an uncomplicated thank you to be certain the NH4+ concentration? (hint: 2 NH4+ ions sort for all and sundry SO4 -- ion)
