Calculate the initial molarity of KNH2 and the molarities of K+, NH3, OH-, and H3O+ in an aqueous solution that contains 0.75 g of KNH2 in 0.255 L of solution. You may ignore the reaction of NH3 with water.
I already found the initial molarity, but I am not sure how to find each species concentration with the information given. Please explain how to solve rather than give me only answers--they are helpful to compare but I want to learn how to do this. I know how to set up the ICE chart, but am unsure what to set my equation equal to.
The equation I am using for my ICE chart is set up as NH2^- + HOH --> NH3+OH^-....I am a little confused where I find the concentration of H3O+
(DrBob, I did see your explanation, and I see the general idea of how to proceed with this, but am stuck)
Trixie--I apologize for not getting back sooner but I looked for Ka NH3 in all of my references and spent about three hours looking on the web but I couldn't find it. I can't believe I've been in the business this many years and I've not seen this before. There are a number of references that show Kb for NH2^- = Kw/Ka, and they calculate Ka as Kw/Kb which just makes Kb for NH2^- regenerate Kb for NH3. I don't think that is correct but I may be mistaken.
A key point to realize is that KNH_2 dissociates completely, and there is an excess of water. If you take that into account, there is no need for Kb or even an ICE chart. The concentration of the products will equal the concentration of the reactants. Hope that clears it up a bit, and hopefully this answer isn't too late to help, haha.
To find the molarities of K+, NH3, OH-, and H3O+ in the solution, you can use the stoichiometry of the balanced chemical equation for the dissociation of KNH2.
First, let's determine the initial molarity of KNH2:
1. Calculate the number of moles of KNH2:
moles of KNH2 = mass of KNH2 / molar mass of KNH2
molar mass of KNH2 = molar mass of K + molar mass of NH2
You can look up the molar masses of K and NH2 in the periodic table.
2. Calculate the initial molarity of KNH2:
initial molarity (M) = moles of KNH2 / volume of solution (in liters)
Convert the given volume of solution (0.255 L) to liters if necessary.
Now, let's determine the molarities of K+, NH3, OH-, and H3O+:
1. Set up an ICE (Initial, Change, Equilibrium) chart for the dissociation of KNH2 into its ions:
NH2- (initial) + HOH (initial) ⇌ NH3 (change) + OH- (change)
2. Identify the stoichiometry of the dissociation reaction:
From the balanced chemical equation, you can see that 1 mole of KNH2 dissociates to give 1 mole of NH3 and 1 mole of OH-.
3. Use the concept of stoichiometry to find the molarities of NH3 and OH-:
If the initial molarity of KNH2 is given (from step 2 in finding initial molarity), it is equal to the initial molarities of NH2- and HOH.
You can assume that the change in NH3 concentration is equal to the initial NH2- concentration (since 1 mole of NH3 is produced for every 1 mole of NH2-).
Similarly, assume that the change in OH- concentration is equal to the initial NH2- concentration.
4. Calculate the molarities of NH3 and OH-:
molarity of NH3 = (initial molarity of NH2-) + (change in NH3 concentration)
molarity of OH- = (initial molarity of NH2-) + (change in OH- concentration)
5. Determine the molarity of K+:
molarity of K+ = initial molarity of KNH2
Regarding the concentration of H3O+, the given information does not provide direct information about its concentration. However, in this case, you can assume that the concentration of H3O+ is equal to the OH- concentration. This assumption is based on the fact that for every OH- ion produced, an H3O+ ion will be formed due to the reaction of NH3 with water.
Therefore, the molarity of H3O+ is equal to the molarity of OH-.
Overall, by following these steps, you should be able to calculate the molarities of K+, NH3, OH-, and H3O+ in the given solution.
You solve that equation I gave you for x. That gives you (OH^-) and (NH3). Convert OH^- to pOH, then to pH, then to H3O^+.
Kb = (Kw/Ka) = (NH3)(OH^-)/(NH2^-)
(Kw/Kb) = (x)(x)/(your molarity -x)
One unknown only. Solve for x. x = (OH^-) = (NH3).