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A solution was obtained by boiling flowers in water - HSC - SSCE Chemistry - Question 19 - 2013 - Paper 1
Question 19
A solution was obtained by boiling flowers in water. After various substances were added to separate samples of the solution, the colour of each was noted. | Substa... show full transcript
Worked Solution & Example Answer:A solution was obtained by boiling flowers in water - HSC - SSCE Chemistry - Question 19 - 2013 - Paper 1
Step 1
For which of the following titrations would it be appropriate to use this solution as an indicator?
Answer
To determine the appropriate titrations for this solution as an indicator, we need to examine the color changes observed when HCl and NaOH were added.
The solution turns bright pink in the presence of HCl, indicating an acidic environment, and bright yellow in the presence of NaOH, indicating a basic environment.
- Option A: HCl(aq) + NH₃(aq) - This interaction involves a weak base (NH₃) with strong acid (HCl), where the indicator can change color, making it appropriate.
- Option B: HCl(aq) + NaOH(aq) - This is a strong acid-strong base titration, where the addition of NaOH to HCl will lead to a neutralization reaction, not utilizing the color change effectively, making it less appropriate.
- Option C: CH₃COOH(aq) + NH₃(aq) - This is a weak acid and weak base titration where the solution will not show a significant color change to indicate the endpoint.
- Option D: CH₃COOH(aq) + NaOH(aq) - A strong base (NaOH) will neutralize the weak acid (CH₃COOH), where again, the indicator may not function effectively as expected.
Consequently, the most appropriate option for using the solution as an indicator is (A) HCl(aq) + NH₃(aq).