Chemistry question help please
Hi, please could someone check if my method for this question is correct?
Here is the question: Sorbic acid (HC6H7O2) is a weak monoprotic acid with Ka = 1.7 x 10-5. Its salt (potassium sorbate) is added to cheese to inhibit the formation of mold. What is the pH of a solution containing 11.25g of potassium sorbate in 1.75dm3 of solution?
Here is my attempt: https://ibb.co/r22L6dV
Thanks!
Here is the question: Sorbic acid (HC6H7O2) is a weak monoprotic acid with Ka = 1.7 x 10-5. Its salt (potassium sorbate) is added to cheese to inhibit the formation of mold. What is the pH of a solution containing 11.25g of potassium sorbate in 1.75dm3 of solution?
Here is my attempt: https://ibb.co/r22L6dV
Thanks!
Reply 1
Original post
by anonymous56754Hi, please could someone check if my method for this question is correct?
Here is the question: Sorbic acid (HC6H7O2) is a weak monoprotic acid with Ka = 1.7 x 10-5. Its salt (potassium sorbate) is added to cheese to inhibit the formation of mold. What is the pH of a solution containing 11.25g of potassium sorbate in 1.75dm3 of solution?
Here is my attempt: https://ibb.co/r22L6dV
Thanks!
Here is the question: Sorbic acid (HC6H7O2) is a weak monoprotic acid with Ka = 1.7 x 10-5. Its salt (potassium sorbate) is added to cheese to inhibit the formation of mold. What is the pH of a solution containing 11.25g of potassium sorbate in 1.75dm3 of solution?
Here is my attempt: https://ibb.co/r22L6dV
Thanks!
Your approach is correct, but not conventional.
In general, this type of problem is solved using Kb = Kw/Ka to arrive at the formula [OH^-] = sqrt Kb × C
Here's my question development:
molar mass(C6H7KO2) = (12 × 6) + (1 × 7) + 39 + (16 × 2) = 150 g/mol.
n = m/M = 11.25 g/150 g/mol = 0.075 mol.
C = n/V = 0.075 mol/1.75 dm^3 = 0.04286 mol/dm^3.
I must determine the hydrolysis constant (Kb).
Kb = Kw/Ka,⬅️ First error corrected here.
Kw = 10^-14 (the ionization constant of water).
Kb = (10^-14)/(1.7 × 10^-5) = 5.88 × 10^-10.⬅️ Second error corrected here.
The hydrolysis reaction is:
C6H7O2^- + H2O <==> HC6H7O2 + OH^-.
The concentration of OH^- ions can be found through:
[OH^-] = sqrt Kb × C.
Substituting:
[OH^-] = sqrt (5.88 × 10^-10) × (0.04286) = sqrt 2.52 × 10^-11 = 5.02 × 10^-6 mol/dm^3.
I will calculate pOH and pH.
pOH = -log10[OH^-] = -log10(5.02 × 10^-6) = 5.30.
pH = 14 - pOH = 14 - 5.30 = 8.70.
PS: I corrected my first version according to TypicalNerd's comments (reply #3 and 5), because I messed up my calculator, I am a bit rusty in using it.
Thanks, TypicalNerd!😀
Bye,
Sandro
(edited 12 months ago)
Reply 2
13
Original post
by NitrotolueneYour approach is correct, but not conventional.
In general, this type of problem is solved using Kb = Kw × Ka to arrive at the formula [OH^-] = sqrt Kb × C
Here's my question development:
molar mass(C6H7KO2) = (12 × 6) + (1 × 7) + 39 + (16 × 2) = 150 g/mol.
n = m/M = 11.25 g/150 g/mol = 0.075 mol.
C = n/V = 0.075 mol/1.75 dm^3 = 0.04286 mol/dm^3.
I must determine the hydrolysis constant (Kb).
Kb = Kw × Ka,
Kw = 10^-14 (the ionization constant of water).
Kb = (1.7 × 10^-5) × (10^-14) = 5.88 × 10^-10.
The hydrolysis reaction is:
C6H7O2^- + H2O <==> HC6H7O2 + OH^-.
The concentration of OH^- ions can be found through:
[OH^-] = sqrt Kb × C.
Substituting:
[OH^-] = sqrt (5.88 × 10^-10) × (0.04286) = sqrt 2.52 × 10^-11 = 1.59 × 10^-6 mol/dm^3.
I will calculate pOH and pH.
pOH = -log10[OH^-] = -log10(1.59 × 10^-6) = 5.80.
pH = 14 - pOH = 14 - 5.80 = 8.20.
Bye,
Sandro
In general, this type of problem is solved using Kb = Kw × Ka to arrive at the formula [OH^-] = sqrt Kb × C
Here's my question development:
molar mass(C6H7KO2) = (12 × 6) + (1 × 7) + 39 + (16 × 2) = 150 g/mol.
n = m/M = 11.25 g/150 g/mol = 0.075 mol.
C = n/V = 0.075 mol/1.75 dm^3 = 0.04286 mol/dm^3.
I must determine the hydrolysis constant (Kb).
Kb = Kw × Ka,
Kw = 10^-14 (the ionization constant of water).
Kb = (1.7 × 10^-5) × (10^-14) = 5.88 × 10^-10.
The hydrolysis reaction is:
C6H7O2^- + H2O <==> HC6H7O2 + OH^-.
The concentration of OH^- ions can be found through:
[OH^-] = sqrt Kb × C.
Substituting:
[OH^-] = sqrt (5.88 × 10^-10) × (0.04286) = sqrt 2.52 × 10^-11 = 1.59 × 10^-6 mol/dm^3.
I will calculate pOH and pH.
pOH = -log10[OH^-] = -log10(1.59 × 10^-6) = 5.80.
pH = 14 - pOH = 14 - 5.80 = 8.20.
Bye,
Hi, thanks for your reply. I got 8.7 but you got 8.2, I understand your method but is my answer still correct?
Reply 3
Original post
by NitrotolueneYour approach is correct, but not conventional.
In general, this type of problem is solved using Kb = Kw × Ka to arrive at the formula [OH^-] = sqrt Kb × C
Here's my question development:
molar mass(C6H7KO2) = (12 × 6) + (1 × 7) + 39 + (16 × 2) = 150 g/mol.
n = m/M = 11.25 g/150 g/mol = 0.075 mol.
C = n/V = 0.075 mol/1.75 dm^3 = 0.04286 mol/dm^3.
I must determine the hydrolysis constant (Kb).
Kb = Kw × Ka,
Kw = 10^-14 (the ionization constant of water).
Kb = (1.7 × 10^-5) × (10^-14) = 5.88 × 10^-10.
The hydrolysis reaction is:
C6H7O2^- + H2O <==> HC6H7O2 + OH^-.
The concentration of OH^- ions can be found through:
[OH^-] = sqrt Kb × C.
Substituting:
[OH^-] = sqrt (5.88 × 10^-10) × (0.04286) = sqrt 2.52 × 10^-11 = 1.59 × 10^-6 mol/dm^3.
I will calculate pOH and pH.
pOH = -log10[OH^-] = -log10(1.59 × 10^-6) = 5.80.
pH = 14 - pOH = 14 - 5.80 = 8.20.
Bye,
Sandro
In general, this type of problem is solved using Kb = Kw × Ka to arrive at the formula [OH^-] = sqrt Kb × C
Here's my question development:
molar mass(C6H7KO2) = (12 × 6) + (1 × 7) + 39 + (16 × 2) = 150 g/mol.
n = m/M = 11.25 g/150 g/mol = 0.075 mol.
C = n/V = 0.075 mol/1.75 dm^3 = 0.04286 mol/dm^3.
I must determine the hydrolysis constant (Kb).
Kb = Kw × Ka,
Kw = 10^-14 (the ionization constant of water).
Kb = (1.7 × 10^-5) × (10^-14) = 5.88 × 10^-10.
The hydrolysis reaction is:
C6H7O2^- + H2O <==> HC6H7O2 + OH^-.
The concentration of OH^- ions can be found through:
[OH^-] = sqrt Kb × C.
Substituting:
[OH^-] = sqrt (5.88 × 10^-10) × (0.04286) = sqrt 2.52 × 10^-11 = 1.59 × 10^-6 mol/dm^3.
I will calculate pOH and pH.
pOH = -log10[OH^-] = -log10(1.59 × 10^-6) = 5.80.
pH = 14 - pOH = 14 - 5.80 = 8.20.
Bye,
Sandro
The correct formula is Kb = Kw/Ka. I can promise you that 1.7 x 10^-5 x 10^-14 ≠ 5.88 x 10^-10, thought 5.88 x 10^-10 is the correct Kb.
Furthermore, sqrt(2.52 x 10^-11) = 5.02 x 10^-6, which if taken to be the hydroxide ion concentration gives an answer of 8.70 as the pH (provided the calculations are completed correctly).
I think the OP’s method and answer are perfectly fine. You can use pOH, but I generally prefer using Kw again, as in my experience of mentoring post-16 level chemistry it results in fewer calculation errors.
Reply 4
Original post
by anonymous56754Hi, thanks for your reply. I got 8.7 but you got 8.2, I understand your method but is my answer still correct?
Take another look at your calculations to make sure they're correct, as there may be some little calculation errors or approximations, but your development of the problem is valid and acceptable.
But why, when you got the [OH^-], did you not go straight to pOH, a conventional solution?
Bye,
Reply 5
Original post
by anonymous56754Hi, thanks for your reply. I got 8.7 but you got 8.2, I understand your method but is my answer still correct?
Your answer is fine. There appears to have been a computation error in Nitrotoluene’s answer, though their method was mostly alright
Reply 6
13
Original post
by TypicalNerdThe correct formula is Kb = Kw/Ka. I can promise you that 1.7 x 10^-5 x 10^-14 ≠ 5.88 x 10^-10, thought 5.88 x 10^-10 is the correct Kb.
Furthermore, sqrt(2.52 x 10^-11) = 5.02 x 10^-6, which if taken to be the hydroxide ion concentration gives an answer of 8.70 as the pH (provided the calculations are completed correctly).
I think the OP’s method and answer are perfectly fine. You can use pOH, but I generally prefer using Kw again, as in my experience of mentoring post-16 level chemistry it results in fewer calculation errors.
Furthermore, sqrt(2.52 x 10^-11) = 5.02 x 10^-6, which if taken to be the hydroxide ion concentration gives an answer of 8.70 as the pH (provided the calculations are completed correctly).
I think the OP’s method and answer are perfectly fine. You can use pOH, but I generally prefer using Kw again, as in my experience of mentoring post-16 level chemistry it results in fewer calculation errors.
Thanks 🙂
Reply 7
13
Original post
by NitrotolueneTake another look at your calculations to make sure they're correct, as there may be some little calculation errors or approximations, but your development of the problem is valid and acceptable.
But why, when you got the [OH^-], did you not go straight to pOH, a conventional solution?
Bye,
Sandro
But why, when you got the [OH^-], did you not go straight to pOH, a conventional solution?
Bye,
Ah, I just prefer doing using Kw idk why it’s just habit I supposed 😁
Quick Reply
Related discussions
- What Would You Like To See In The Chemistry Forum?
- Eligibility for bursaries
- Entry test Medical University Pleven (2021)
- How to get better at physical chemisrry
- OCR A A-Level Chemistry resources
- I’m resiting a level chemistry and a level biology.HELP PLEASE!!
- Can I apply for pharmacy with an AAB
- A level tips.
- Courses
- A level Biology and chemistry flashcards!
- Chemistry a level help please urgent
- chemistry a level aqa year 12
- Uni
- Read if you’ve finished A level chem with A/A*
- Chemistry A level Help!!!
- Help with choosing medical schools
- Biochem or Biomed: Which is better for me?
- Chemistry a level mocks
- A Levels
- Chemistry a level
Latest
Trending
Last reply 2 days ago
AQA A level chemistry synthetic routes map - what is the best one to memorise?Last reply 3 months ago
AQA Chemistry Electrode Potentials Conventional Representation QuestionLast reply 5 months ago
OCR B Salters A Level Chemistry 2025 Advanced ArticleLast reply 6 months ago
Help with this chemistry question? (A2, acid-base equilibria)Articles for you
How The Student Room is moderated
To keep The Student Room safe for everyone, we moderate posts that are added to the site.
