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What is the volume percent of water in 14.8 M ammonium hydroxide?
This seems simple enough, but the solution eludes me. I'm trying to determine how much water is added through the ammonium hydroxide when making a 1.87 M NH3 solution of 100ml using 14.8 M stock Ammonium Hydroxide (25-28%). Thanks for the help.
I can calculate the 0.126351 L of Ammonium necessary to get a 1.87 Molarity but I'm trying to find the volume percent of that 0.126351 L which is composed of the H20, and not the ammonia. I'm getting about 0.04L or 38 vol% but I'm not sure since ammonia is a gas instead of a solid and I don't know how that affects the composition and volume percent such as in response to temperature.
2 Answers
- DavidBLv 74 years ago
mass percent is easy volume percent is not. I don't think you can calculate the volume percent. Ammonia is a gas and a weak base in solution - in solution it is still mostly NH3 though sometimes labelled NH4OH. You can't calculate a volume for NH4OH and ammonia does not stay as a gas. I would go with mass percent.
- hcbiochemLv 74 years ago
Your question is easy to answer without knowing how much water is present in the original solution. You are simply doing a dilution calculation which uses the formula:
M1V1 = M2V2
(14.8 M) (V1) = 1.87 M (100 mL)
V1 = 12.6 mL
So, by taking 12.6 mL of the 14.8 M NH4OH solution and adding water to a final volume of 100 mL, the resulting solution will have a concentration of 1.87 M.
(As an aside, ammonium hydroxide is somewhat of a misnomer. It is just a solution of NH3 in water.)