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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
The solution demonstrates distinct color changes, indicating approximate pH ranges:
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HCl(aq) + NH₃(aq): This reaction occurs between a strong acid (HCl) and a weak base (NH₃), which will produce an acidic solution. Hence, it utilizes the color change from bright pink to pale yellow, this would be suitable for the indicator.
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HCl(aq) + NaOH(aq): This titration is between a strong acid and a strong base. The solution would change colors but might not give a clear endpoint as it tends to neutralize.
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CH₃COOH(aq) + NH₃(aq): Here we have a weak acid and a weak base, which would not provide a clear enough change of the indicator.
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CH₃COOH(aq) + NaOH(aq): This involves a weak acid and a strong base; the solution would respond well to an indicator similar to that of HCl and NaOH, but may not provide the same range of color change.
Conclusion: Option (A) HCl(aq) + NH₃(aq) is the most appropriate titration or any case where significant pH variations leading to color change occur.