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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)