Ammonia (NH3) is a weak base with a Kb= 1.8 X 10^-5 and hydrochloric acid (HCl) is a strong acid. What is the pH of a mixture made by adding 6.61 grams of NH4Cl to a 412 ml of a 0.180M solution of NH3?
Use the Henderson-Hasselbalch equation.
Post your work if you get stuck.
DrBob222, the pka value would be 9.26,. but how would you figure out the concentrations of the acid and conjugate base (M). You would have 0.124 mol NH4Cl and 0.07416 mol NH3, but how do you get them to concentrations. I know I could divide each by the total volume, but the problem does not give that info. Help Please
Re-read the problem. Looks like 412 mL to me. And both are in the same 412 mL. Actually, you don't need to know the volume for it cancels.
Note M = moles/L.
So M NH3 = 0.0742/0.412 L = ??
M NH4Cl = 0.124/0.412 L = ??
BUT since you are dividing base/acid that is
pH = pKa + log (b/a)
pH = pKa + log (moles b/0.412/moles a/0.412) and the 0.412 cancels leaving just
pH = pKa + log (mols b/mols a).
Thanks for poiting that out. I got a pH of 9.03. My only question is why the pH isn't acidic if it is a strong acid mixed with a weak base?
It isn't a strong acid mixed with a weak base. That is the salt. The salt, NH4Cl, IS the salt of a strong acid and a weak base, as you point out; however, you have that salt mixed with AMMONIA which is more of the weak base. And a weak base of ammonia is in excess. Furthermore, a weak base and it's salt makes a BASIC buffer.
The water solution of NH4Cl WOULD be acidic but adding ammonia to it makes it basic.
To find the pH of the mixture, we need to determine the concentration of the resulting solution after adding NH4Cl to the NH3 solution.
Let's start by calculating the number of moles of NH4Cl added to the solution:
1. Calculate the molar mass of NH4Cl:
NH4Cl = (1 × 14.01 g/mol) + (4 × 1.01 g/mol) + 35.45 g/mol
= 53.49 g/mol
2. Convert the mass of NH4Cl to moles:
moles of NH4Cl = 6.61 g / 53.49 g/mol
= 0.1236 mol
Next, we need to find the initial moles of NH3 present in the solution:
1. Calculate the moles of NH3:
moles of NH3 = concentration of NH3 × volume of solution (in liters)
volume of solution = 412 ml = 412/1000 L = 0.412 L
moles of NH3 = 0.180 M × 0.412 L
= 0.07416 mol
Now, let's calculate the moles of NH3 remaining after the reaction with NH4Cl:
1. NH3 reacts with NH4Cl to form NH4+ and OH-.
So, the moles of NH4Cl will react with an equal number of moles of NH3, forming moles of NH4+.
2. Since NH3 and NH4+ exist in a 1:1 ratio, the moles of NH4+ formed will be equal to moles of NH4Cl added.
So, moles of remaining NH3 = initial moles of NH3 - moles of NH4Cl added
= 0.07416 mol - 0.1236 mol
= -0.04944 mol
However, it doesn't make sense to have a negative amount of NH3 remaining. This indicates that all of the NH3 has reacted, and we have an excess of NH4Cl.
Now, let's find the concentration of NH4+ formed from the reaction:
1. The volume of the solution remains the same, so the concentration of NH4+ will be:
concentration of NH4+ = moles of NH4+ formed / volume of solution
2. Since moles of NH4+ formed = moles of NH4Cl added, the concentration of NH4+ will be:
concentration of NH4+ = 0.1236 mol / 0.412 L
= 0.300 mol/L
At this point, we know the concentration of NH4+ in the solution. However, we need to calculate the concentration of OH- to determine the pH.
Since NH4+ is the conjugate acid of NH3, we can use the Kb value to find the concentration of OH-:
1. Kb = [NH4+][OH-] / [NH3]
Let x be the concentration of OH-.
2. Substitute the known values:
1.8 × 10^-5 = (0.300 mol/L) × x / (0.07416 mol/L)
3. Solve for x:
x = (1.8 × 10^-5 × 0.07416 mol/L) / 0.300 mol/L
= 4.424 × 10^-6 mol/L
Now, we can calculate the pOH of the solution:
1. pOH = -log10[OH-]
pOH = -log10(4.424 × 10^-6)
≈ 5.354
Finally, let's find the pH of the solution using the pOH:
1. pH + pOH = 14
pH + 5.354 = 14
2. Solve for pH:
pH ≈ 14 - 5.354
≈ 8.646
Therefore, the pH of the mixture is approximately 8.646.