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. Thanks!
To find the molarities of each species in the solution, we need to calculate the number of moles of each component present and then divide by the total volume of the solution.
1. Calculate the number of moles of KNH2:
Given mass = 0.75 g
Molar mass of KNH2 = molar mass of K (39.10 g/mol) + molar mass of NH2 (15.03 g/mol) = 54.13 g/mol
Number of moles = given mass / molar mass = 0.75 g / 54.13 g/mol
2. Calculate the initial molarity of KNH2:
Molarity (M) = number of moles / volume of solution in liters
Initial molarity of KNH2 = number of moles / 0.255 L
Next, we need to consider the dissociation reaction of KNH2 in water:
KNH2 → K+ + NH2-
3. Determine how many moles of K+ and NH2- are produced:
Since KNH2 dissociates into one K+ ion and one NH2- ion, the number of moles of K+ and NH2- are the same as the number of moles of KNH2.
4. Calculate the molarities of K+ and NH2-:
Molarity of K+ = moles of K+ / volume of solution in liters
Molarity of NH2- = moles of NH2- / volume of solution in liters
Now, let's move on to finding the molarities of OH- and H3O+ ions. Since we are ignoring the reaction of NH3 with water, we can assume that all the NH2- ions react with water to produce OH- and NH3:
NH2- + H2O → NH3 + OH-
5. Calculate the number of moles of NH2-:
Number of moles of NH2- = moles of KNH2
6. Calculate the number of moles of NH3 and OH- produced:
Since NH2- reacts with water to produce one NH3 molecule and one OH- ion, the moles of NH3 and OH- produced are the same as the moles of NH2-.
7. Calculate the molarities of NH3 and OH-:
Molarity of NH3 = moles of NH3 / volume of solution in liters
Molarity of OH- = moles of OH- / volume of solution in liters
Lastly, we know that in a neutral solution, the concentration of H3O+ ions is equal to the concentration of OH- ions:
H3O+ + OH- → 2H2O
8. Calculate the molarity of H3O+:
Molarity of H3O+ = Molarity of OH-
By following these steps, you can calculate the molarities of K+, NH3, OH-, and H3O+ in the given solution.
To find the molarities of K+, NH3, OH-, and H3O+ in the aqueous solution, we need to consider the stoichiometry of the compound KNH2.
1. Calculate the number of moles of KNH2:
To find the number of moles, we use the formula:
Moles = mass (g) / molar mass (g/mol)
Given: mass of KNH2 = 0.75 g
The molar mass of KNH2 = (atomic mass of K) + (atomic mass of N) + 2 * (atomic mass of H)
Now, looking up the atomic masses:
atomic mass of K = 39.10 g/mol
atomic mass of N = 14.01 g/mol
atomic mass of H = 1.01 g/mol
Molar mass of KNH2 = 39.10 + 14.01 + 2 * 1.01 = 55.13 g/mol
So, Moles of KNH2 = 0.75 g / 55.13 g/mol
2. Calculate the initial molarity of KNH2:
Molarity (M) is defined as moles of solute per liter of solution.
Initial Molarity of KNH2 = moles of KNH2 / volume of solution (in liters)
Given: volume of solution = 0.255 L
Initial Molarity of KNH2 = Moles of KNH2 / 0.255 L
3. Calculate the concentration of K+ and NH3:
Since KNH2 dissociates into K+ and NH2-, the concentration of K+ and NH3 will be the same as the initial Molarity of KNH2 since there is no reaction with water.
4. Calculate the concentration of OH- and H3O+:
As mentioned in the question, we need to ignore the reaction of NH3 with water. This means that NH3 does not produce OH- or H3O+. Therefore, the concentrations of OH- and H3O+ will be zero.
So, to recap:
- The initial Molarity of KNH2 is calculated using the moles of KNH2 divided by the volume of the solution.
- The molarities of K+ and NH3 are equal to the initial Molarity of KNH2.
- The molarities of OH- and H3O+ are both zero.
Remember to use the appropriate units (moles, liters) when performing calculations to determine the concentrations.
NH2^- + HOH ==> NH3 + OH^-
Set up ICE chart with NH3 = x and OH^- = x so NH2^- = your molarity -x
Solve for x.
Kb = (Kw/Ka) = (NH3)(OH^-)/(NH2^-)
That gives you NH3, OH, you have NH2^- and K^+ is same as NH2^-.