A 10.0 mL solution of 0.300 M NH3 is titrated with a 0.100 M HCl solution. Calculate the pH after the addition of 20.0 mL of the HCl solution.
(potentially useful info: Ka of NH4+ = 5.6 x 10−10)
millimoles NH3 = 10 x 0.3 = 3
mmols HCl = 20 x 0.1 = 2
NH3 + HCl => NH4Cl
So you have 2 mmols NH4^+
and 1 mmol NH3.
pH = pKa + log (base)/(acid)
Thanks, I got the answer 8.95. So why do I not need to divide the mmols by the total volume?
Are you ready for the full answer? It will take a bit of reading.
The Henderson-Hasselbalch equation is
pH = pKa + log(base)/(acid)
With regard to the (base) and (acid), that means CONCENTRATION of base and CONCENTRATION of acid. Technically, then, one must divide millimoles/milliliter to get concn. But if you will put in the numbers (let's use made up numbers of 2 mmols for base and 3 mmols for acid and a volume of 10 mL. So this log ratio will come up with
(2/10)/(3/10) but notice that the 10 cancels and you have simply 2/3. Since it is the SAME volume in the solution for both acid and base you will note that the volume will ALWAYS cancel so you can simply use mmols base/mmols acid. When I taught this years ago I always counted off a few points because I told the students it was (concn/concn) not (mmoles/mmoles) and never mind that the answer was the same. So if you want to be technically correct you will use (mmols/mL)/(mmols/mL). If you want to do it quick and dirt you can simply use mmols because the answer will be the same. In fact, when I taught this I NEVER actually divided mmols/mL and plugged in that number; I always wrote mmols and mL, canceled the mL and went from there. Fast forward 25 years: I note some of the profs now use mmols directly and don't insist on the concn stage. I think it's ok to use that procedure as long as we understand what we are doing. If I were a student in a class these days I would ask the prof how s/he wanted it done on exams.
Thank you for the explanation!
To calculate the pH after the addition of the HCl solution, let's break down the process step by step.
1. Determine the moles of NH3 initially present:
Moles of NH3 = volume (in liters) x molarity
Moles of NH3 = 10.0 mL / 1000 mL/L x 0.300 M
Moles of NH3 = 0.003 moles
2. Determine the moles of HCl added:
Moles of HCl = volume (in liters) x molarity
Moles of HCl = 20.0 mL / 1000 mL/L x 0.100 M
Moles of HCl = 0.002 moles
3. Determine the moles of NH4+ formed:
The reaction between NH3 and HCl is as follows:
NH3 + HCl -> NH4Cl
The balanced equation shows a 1:1 mole ratio. Therefore, moles of NH4+ = moles of HCl added = 0.002 moles
4. Calculate the concentration of NH4+ after the reaction:
The volume of the solution after adding the HCl solution is the sum of the initial NH3 volume and the HCl volume:
Total volume = 10.0 mL + 20.0 mL = 30.0 mL
Convert the volume to liters:
Total volume = 30.0 mL / 1000 mL/L = 0.030 L
Concentration of NH4+ = moles of NH4+ / total volume
Concentration of NH4+ = 0.002 moles / 0.030 L = 0.067 M
5. Convert the concentration of NH4+ into NH4+ and NH3:
Since NH4+ is a weak acid, it will partially dissociate according to the given Ka value: Ka = 5.6 x 10^-10
NH4+ <--> NH3 + H+
Let's assume x M of NH4+ dissociates.
At equilibrium, the concentration of NH4+ is 0.067 - x M.
The concentration of H+ is also x M.
Using the equation for the Ka expression:
Ka = [NH3][H+] / [NH4+]
5.6 x 10^-10 = x * x / (0.067 - x)
Since x is small compared to 0.067, we can disregard x when subtracting from 0.067:
5.6 x 10^-10 = x * x / 0.067
Solve the equation for x by rearranging and solving the quadratic equation:
x^2 = 5.6 x 10^-10 * 0.067
x^2 = 3.752 x 10^-11
x = sqrt(3.752 x 10^-11)
x = 6.12 x 10^-6 M
6. Calculate the concentration of H+:
Since H+ concentration equals x, the concentration of H+ is 6.12 x 10^-6 M.
7. Calculate the pH:
The pH is calculated using the equation: pH = -log10[H+]
pH = -log10(6.12 x 10^-6)
pH ≈ 5.21
Therefore, the pH after the addition of 20.0 mL of the HCl solution is approximately 5.21.