To a 100 mL volumetric flask were added 5.00 mL of 6.0 M NH3 , 14.5g of solid ammonium iodide, and 40.0 mL of 0.50 M NaOH. The flask was filled to the calibration mark with water.
(a) What were the molar concentrations of hydrogen ion, hydroxide ion, and ammonium ion in the resultant solution? K = 1.8 x 10-5 for NH3
what do you do with the solid in terms of concentration? How do you determine the dominant equation?
I think the first thing you need to do is to write ionization equations and calculate the molarity of each of the components. For the solid, you divide the grams by molar mass NH4I to determine mols and divide that by vol (in Liters) to convert that to molarity. If I didn't goof NH4I is 1 M.
Then remembering NaOH is a strong base and ionizes 100% and NH3 is a weak base with a Kb (relatively small) PLUS the NH4I will act as a common ion (the NH4 from NH4I is common to the NH4^+ produced by NH3 + HOH ==> NH4^+ + OH^-
The majority of the NH4^+ will come from the NH4I with a small amount from NH3 reaction with water. You may want to calculate that (or at least estimate it) to see if the amount is significant. The majority of the OH^- will come from NaOH (also NaOH is a common ion to the NH3 reaction, also, which reduces the contribution of the NH3 reaction both to the NH4^+ as well as to the OH^-). I didn't go through the calculation but I suspect the NH4^+ and the OH^- from the NH3 + HOH reaction will be VERY small, and possibly negligible.
To determine the molar concentrations of hydrogen ion, hydroxide ion, and ammonium ion in the solution, we need to consider the reactions that occur between the different species in the flask.
Let's break down the steps to solve this problem:
1. Determine the dominant equation: We need to consider the reactions that can occur between the different species present in the flask. In this case, we have three reactants: NH3, ammonium iodide (NH4I), and NaOH.
First, let's consider NH3. Since NH3 is a weak base, it can react with water to generate hydroxide ions (OH-) and ammonium ions (NH4+):
NH3 + H2O ⇌ NH4+ + OH-
Next, let's consider NH4I. This compound can dissociate in water to release ammonium ions (NH4+) and iodide ions (I-):
NH4I ⇌ NH4+ + I-
Finally, let's consider NaOH. This compound is a strong base and dissociates completely in water to form sodium ions (Na+) and hydroxide ions (OH-):
NaOH → Na+ + OH-
2. Assess solid composition: In this case, we have solid ammonium iodide (NH4I). Since it is not in solution, it does not contribute to the molar concentrations of the ions in the solution. We will exclude it from further calculations.
3. Determine the moles: To calculate the molar concentrations of the ions, we need to determine the moles of each species present in the solution.
First, let's determine the moles of NH3:
Moles of NH3 = volume (L) × concentration (mol/L)
= (5.00 mL ÷ 1000 mL/L) × 6.0 M
= 0.0300 mol
Next, let's determine the moles of NaOH:
Moles of NaOH = volume (L) × concentration (mol/L)
= (40.0 mL ÷ 1000 mL/L) × 0.50 M
= 0.0200 mol
4. Determine the final volume: Since the flask was filled to the calibration mark with water, the total volume of the solution is 100 mL.
5. Calculate the molar concentrations: Now that we have the moles of NH3 and NaOH, as well as the total volume of the solution, we can calculate the molar concentrations of the ions.
The molar concentration of hydroxide ions (OH-) is equal to the concentration of NaOH added because NaOH dissociates completely:
[OH-] = (moles of NaOH) / (volume of solution in liters)
= 0.0200 mol / 0.100 L
= 0.20 M
The molar concentration of ammonium ions (NH4+) can be determined using the equilibrium expression and the known equilibrium constant (K) for NH3:
[NH4+] = (K × [OH-]) / [NH3]
= (1.8 × 10^-5) × (0.20 M) / (0.0300 M)
= 0.12 M
Finally, the molar concentration of hydrogen ions (H+) can be calculated using the fact that water is a neutral solution and the concentration of hydroxide ions (OH-):
[H+] = Kw / [OH-]
= (1.0 × 10^-14 M^2) / 0.20 M
= 5.0 × 10^-14 M
So, the molar concentration of hydrogen ions (H+) is 5.0 × 10^-14 M, the molar concentration of hydroxide ions (OH-) is 0.20 M, and the molar concentration of ammonium ions (NH4+) is 0.12 M in the resultant solution.