Titration calculations and Ksp calculations
-
Hi all, need help in checking approach for these 2 questions.
Titration Question
Calculate the pH of the solution when 20.0 cm^3 of 0.1 M of NaOH was added to 20.0 cm^3 of 0.1 M of H2SO4.
Assume the first ionisation is complete while the Ka for 2nd stage of ionisation is =0.011M
Approach
1st stage: NaOH + H2SO4 --> NaHSO4 + H2O
Moles of HSO4- in solution = 0.002 mol
[HSO4- ]= 0.002 mol/ 0.04 dm^3 = 0.05M
Assuming little ionisation of HSO4-,
[H+] = sqrt ( Ka x [HSO4-]) = sqrt (0.011 x 0.05M) = 0.002345
pH = -log (0.002345) = 1.63
-
Ksp Question
i) Calculate the [OH-] of 0.1 M of NH3 solution. (Kb = 1 x 10-5 M)
[OH-] = sqrt ( Kb x [NH3]) = sqrt (0.00001 x 0.1) =0.001 M
ii) Calculate the ionic product of Ca(OH)2 when 0.1 M of CaCl2 is added to equal volumes of NH3 solution
[Ca2+] = 0.1/2 = 0.05 M
[OH-] = ( Kb x [NH3])= sqrt [ 0.00001 x (0.1)/2]= 0.000707 M
as opposed to just dividing [OH-] from part i) by 2.
[Ca2+][OH-]^2 = (0.05) (0.000707)^2 = 2.5 x 10-8
-
Yes, both your workings are essentially correct. Just a couple of comments.
------------------------------
The 2nd proton dissociation of H2SO4 is significant. Afterall, your own calculations indicate the pH to be less than 2. Therefore, it's technically inaccurate* to say "Assuming little ionisation of HSO4-", rather, what you meant was "Assuming it's mathematically valid to approximate equilibrium molarity of HSO4- back to its initial molarity". Mathematically, such an approximation would be valid only if the initial molarity was at least 1000x larger than the Kc (eg. Ka) value, which it's not, in this case. Which means you cannot use approximation, and have to solve the quadratic equation (eg. via a graphing calculator). However, Cambridge has specified that for 'A' level purposes, they will ensure that in their questions, the initial molarity is always at least 1000x larger than the Kc value, so that the approximation is mathematically valid.
(* there's a 2nd reason why your statement is technically inaccurate. "Ionization" (unless you specify "2nd ionization energy", etc) implies to generate an ion from a non-ion, hence to say "H2SO4 ionizes to HSO4-" is fine, but to say "HSO4- ionizes to SO4 2-" is technically inaccurate. Hence, a better phrasing for Bronsted-Lowry acids would be "proton dissociation" rather than "ionization", and a better phrasing for Bronsted-Lowry bases would be "hydrolysis" rather than "ionization" or "dissociation").
------------------------------------
Yes, your method for determining molarity of OH- is correct : you have to re-use the Kb formula in the new volume, instead of just dividing the initial molarity by 2. This is because the position of equilibrium shifts (as predicted by Le Chatelier's principle) when more water is added, providing for a greater availability of protons for the NH3 base to abstract, hence generating more OH- ions in the new equilibrium compared to simply dividing the previous molarity of OH- ions by half, ie. 0.0007 > (0.001)/2, but because the new molarity of NH3 is itself halved in the newly doubled volume, hence the new molarity of OH- is still less than the previous molarity, 0.0007 < 0.001.
-
Hi Ultima,
Thanks for taking the time.
You're right on the limitations of the approximation approach.
So I did both the approximation (pH = 1.63) and quadratic method (pH = 3.13) and posted only the approximation approach.
The question was actually from a JC worksheet, hence my doubts was if it was deviating from the requirements of not needing quadratic equations.
If I could squeeze in another chemical equilibrium question. This was described to me verbally, and so I am not sure if it contains sufficient info. Still stumped after thinking for while.
Chemical Equilibrium
Sulfur dioxide and oxygen reacts to form sulfur trioxide
2SO2 + O2 <--> 2SO3 Kp = 0.13 atm
Some SO3 was added to a container and equilibirum was reached.
If P is the total pressure at any time, show that intial pressure of SO3 is (P minus Partial pressure of O2)
-
-
Originally posted by atomos:
Hi Ultima,
Thanks for taking the time.
You're right on the limitations of the approximation approach.
So I did both the approximation (pH = 1.63) and quadratic method (pH = 3.13) and posted only the approximation approach.
The question was actually from a JC worksheet, hence my doubts was if it was deviating from the requirements of not needing quadratic equations.
If I could squeeze in another chemical equilibrium question. This was described to me verbally, and so I am not sure if it contains sufficient info. Still stumped after thinking for while.
Chemical Equilibrium
Sulfur dioxide and oxygen reacts to form sulfur trioxide
2SO2 + O2 <--> 2SO3 Kp = 0.13 atm
Some SO3 was added to a container and equilibirum was reached.
If P is the total pressure at any time, show that intial pressure of SO3 is (P minus Partial pressure of O2)
The flawed question both contradicts and proves itself. It implies that at initial, only SO3 was present. Hence the initial pressure of SO3 = P - 0 = P, since P is (ridiculously) defined as total pressure at any time, hence the question proves itself and there's no need for any working.I'd advise you to advise your students not to bother chasing the dodgy and the dubious. Many JC prelim paper questions are so problematic (every year, there will always be many JC teachers who disagree with, openly sneer at and even outright ridicule the prelim paper questions and/or mark schemes from other JCs, and instruct their students to ignore the question and/or mark scheme) that I routinely tell my students to skip many of such questions entirely.
I prefer to set my own questions, and/or use more reliable quality sources such as practice questions from various University textbooks (which I use for teaching my BedokFunland JC students).
Be discriminating when encountering dubious, dodgy questions, in your (students') practice work.
-
That was my original suspicions at first, it was part of a question (provided no other info was omitted) they had to answer for a test.
For one thing, it didn't exactly specify if O2 and SO2 was already present in the first place, so I wasnt sure if it that should be taken in consideration...or if it even matters at all.
Still got alot to learn in sieving out dubious questions.
Thanks for your input!