A sample of 8.00 L of NH3 (ammonia) gas at 22 degree Celsius and 735 torr is bubbled into a 0.500 L solution of 0.400 M (hydrochloric acid).
The kb value for NH3 is 1.8 x 10^-5.
Assuming all the dissolves and that the volume of the solution remains at 0.500L , calculate the pH of the resulting solution.
To calculate the pH of the resulting solution, we need to find the concentration of the resulting NH4+ ions (ammonium ions) in the solution. From there, we can use the equation for Kb to find the concentration of hydroxide ions (OH-) and then use that to calculate the pOH and finally, pH.
Let's break down the steps to get the solution:
Step 1: Convert the temperature from Celsius to Kelvin.
Given: 22 °C
To convert to Kelvin, add 273.15 to the Celsius value.
Temperature in Kelvin (K) = 22 + 273.15 = 295.15 K
Step 2: Calculate the partial pressure of NH3 before it dissolves.
Given: NH3 at 735 torr
Convert the pressure from torr to atm by dividing by 760 (since 1 atm = 760 torr).
Partial pressure of NH3 (P) = 735/760 = 0.967 atm
Step 3: Calculate the moles of NH3.
To calculate the moles of NH3, we need to use the Ideal Gas Law Equation: PV = nRT
Given: P = 0.967 atm, V = 8.00 L, R = 0.0821 L·atm/(mol·K), T = 295.15 K
Rearranging the equation, we get: n = PV / RT
Moles of NH3 (n) = (0.967 atm * 8.00 L) / (0.0821 L·atm/(mol·K) * 295.15 K)
Now, we have the moles of NH3.
Step 4: Calculate the moles of NH4+ ions produced.
Since NH3 is a base, it reacts with HCl to form NH4+ ions.
The balanced chemical equation for this reaction is:
NH3 + HCl → NH4+ + Cl-
We start with 0.400 M (0.500 L) of HCl in solution.
Moles of NH4+ ions produced = (0.400 moles/L * 0.500 L)
Step 5: Calculate the concentration of NH4+ ions in the final solution.
Given the volume of the solution remained at 0.500 L, and moles of NH4+ ions produced from the previous step, you can calculate the concentration of NH4+ ions (ammonium ions) in the solution by dividing the moles by the new volume.
Concentration of NH4+ ions = (moles of NH4+ ions) / (0.500 L)
Step 6: Calculate the concentration of OH- ions using Kb.
Given: Kb for NH3 = 1.8 × 10^-5
The Kb expression for NH3 is:
Kb = ([NH4+][OH-]) / [NH3]
We have the concentration of NH4+ ions from the previous step. Let's assume the concentration of OH- ions is x M.
Plug in the values to the Kb expression:
1.8 × 10^-5 = (concentration of NH4+ ions) × x / (concentration of NH3)
Rearrange the equation to solve for x (concentration of OH- ions):
x = (1.8 × 10^-5) × (concentration of NH3) / (concentration of NH4+ ions)
Step 7: Calculate pOH.
pOH is the negative logarithm of the hydroxide ion concentration. We have the concentration of OH- ions from the previous step.
pOH = -log10(concentration of OH- ions)
Step 8: Calculate pH.
pH is the negative logarithm of the hydrogen ion concentration. Use the relation: pH + pOH = 14
pH = 14 - pOH
Plug the calculated pOH value into the equation to find pH.
That's it! You can follow these steps to calculate the pH of the resulting solution.
Use PV = nRT to determine n for NH3 at the conditions listed. Write the equation for NH3 + HCl==>
the prepare an ICE chart. I suspect, but I've not worked the problem, that there will be an excess of NH3 for the reaction which will lead to a solution of NH3 and NH4Cl. Use the Henderson-Hasselbalch equation to solve for pH. If my assumption is wrong about being left with a buffer solution, another approach must be used.