Asked by Samy
Consider the dissociation of ammonia in water at equilibrium:
NH3 + H2O ↔ NH4+ + OH–
You start with 0.05 moles of ammonia in 500 mL of water. The equilibrium constant Keq is 1.8 × 10–5. What is the pH of this solution at equilibrium?
NH3 + H2O ↔ NH4+ + OH–
You start with 0.05 moles of ammonia in 500 mL of water. The equilibrium constant Keq is 1.8 × 10–5. What is the pH of this solution at equilibrium?
Answers
Answered by
Samy
Still can't figure this one out after 1 hr of trying
Answered by
DrBob222
NH3 + H2O ↔ NH4+ + OH–
You start with 0.05 moles of ammonia in 500 mL of water.
I think I've done one for you almost like this. First the (NH3) = 0.05 mols/0.500 L = 0.1 M
............NH3 + H2O ↔ NH4+ OH^-
I............0.1M..................0...........0
C..............-x.................... x...........x
E..........0.1-x.....................x............x
Keq = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
1.8E-5 = (x)(x)/(0.1-x)
Solve for x and that gives you (OH^-)
You can calculate pOH from that then go to pH.
Almost all of these ICE charts look the same. Same reasoning. Can't go wrong.
You start with 0.05 moles of ammonia in 500 mL of water.
I think I've done one for you almost like this. First the (NH3) = 0.05 mols/0.500 L = 0.1 M
............NH3 + H2O ↔ NH4+ OH^-
I............0.1M..................0...........0
C..............-x.................... x...........x
E..........0.1-x.....................x............x
Keq = 1.8E-5 = (NH4^+)(OH^-)/(NH3)
1.8E-5 = (x)(x)/(0.1-x)
Solve for x and that gives you (OH^-)
You can calculate pOH from that then go to pH.
Almost all of these ICE charts look the same. Same reasoning. Can't go wrong.
Answered by
Samy
Never heard of the ICE chart can you do it another way so its more clear?
Answered by
Chill
Hey Samy! So the ICE charts are usually what are used for these types of problems so you would prob want to get used to seeing them.
"Basically "I" stands for the initial concentrations (or pressures) for each species in the reaction mixture.
"C" represents the change in the concentrations (or pressures) for each species as the system moves towards equilibrium.
"E" represents the equilibrium concentrations (or pressures) of each species when the system is in a state of equilibrium."
This info is from chem.purdue.edu, in case this is still confusing. It gives you great examples too to practice! Best of luck!
"Basically "I" stands for the initial concentrations (or pressures) for each species in the reaction mixture.
"C" represents the change in the concentrations (or pressures) for each species as the system moves towards equilibrium.
"E" represents the equilibrium concentrations (or pressures) of each species when the system is in a state of equilibrium."
This info is from chem.purdue.edu, in case this is still confusing. It gives you great examples too to practice! Best of luck!
Answered by
Samy
ok will def check that out and can you finish answering this i posted it somewhere else on this web but Dr didnt get back to it and i need it asap
"A calorimeter contains 500 g of water at 25°C. You place a cold pack
containing 100 g of crystalline ammonium nitrate inside the calorimeter. When the ammonium nitrate finishes dissolving, the temperature of the water inside the calorimeter is 9.3°C. The specific heat of water is 4.18 J/g–°C. What is the enthalpy of fusion (ΔHf) of the ammonium nitrate? (Show your work.) Where necessary, use q = mHf."
"A calorimeter contains 500 g of water at 25°C. You place a cold pack
containing 100 g of crystalline ammonium nitrate inside the calorimeter. When the ammonium nitrate finishes dissolving, the temperature of the water inside the calorimeter is 9.3°C. The specific heat of water is 4.18 J/g–°C. What is the enthalpy of fusion (ΔHf) of the ammonium nitrate? (Show your work.) Where necessary, use q = mHf."
Answered by
DrBob222
I can not believe you haven't heard of the ICE chart. Chill described it very well. Where is this course coming from? Don't you have a text? If you don't have a basic understanding of how the ICE chart system works you have a really hard road ahead. Almost any problem I've heard of for Keq, Ksp, K you name it, starts with writing down the initial concentrations followed by the change and then the equilibrium concentrations. You can't do ANYTHING with the K value until you know what to substitute into the K expression.
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