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?
Still can't figure this one out after 1 hr of trying
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!
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."
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.
To determine the pH of the solution at equilibrium, we need to calculate the concentration of H+ ions, which is related to pH.
From the given equilibrium reaction, NH3 + H2O ↔ NH4+ + OH–, we can see that the concentration of NH4+ ions is equal to the concentration of H+ ions, as they are formed in equal amounts during the dissociation of ammonia.
Let's denote the concentration of NH4+ ions as [NH4+] and the concentration of H+ ions as [H+]. At equilibrium, these concentrations will be the same.
First, we need to calculate the concentration of NH4+ ions at equilibrium:
Given:
Initial moles of NH3 = 0.05 moles
Total volume of solution = 500 mL = 0.5 L
To calculate the concentration of NH4+ ions, we need to determine the moles of NH4+ ions in the equilibrium state. Since NH4+ ions are formed in a 1:1 stoichiometric ratio with NH3, the moles of NH4+ ions will also be 0.05 moles.
Now, we convert the moles of NH4+ ions to concentration by dividing by the total volume of the solution:
[NH4+] = 0.05 moles / 0.5 L = 0.1 M
Since the concentration of [NH4+] is equal to the concentration of [H+], we have:
[H+] = 0.1 M
To calculate the pH, we need to take the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]
Substituting the concentration we calculated:
pH = -log(0.1) = -(-1) = 1
Therefore, the pH of the solution at equilibrium is 1.
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.