Question 1 (1 point) For the KHP and NaOH titration, the endpoint is reached whe
ID: 551616 • Letter: Q
Question
Question 1 (1 point) For the KHP and NaOH titration, the endpoint is reached when a very faint pink colour lasts for at least 30 seconds. If your solution ends up a very dark pink colour, will your calculated concentration of NaOH be lower or higher than the true value? Higher because more volume of NaOH is being used to titrate the moles of KHIP No effect because the moles of KHP titrated does not change. Higher because more volume of KHP is being used to titrate the moles of NaOH. Lower because more volume of NaOH is being used to titrate the moles of KHP Lower because more volume of KHP is being used to titrate the moles of NaOHExplanation / Answer
Ans. The indicator used in this acid-base titration with characteristic faint pink color as titration end point is phenolphthalein. The indicator is colorless in acidic as well as neutral solution. The indicator turn pink only when the pH reaches 8.2 or above. And, greater is the amount of extra base added (upto pH 10.0), greater would be the intensity of pink color.
Note that for pink color to occur at endpoint, the acid (KHP) must be taken in the flask and NaOH must be in the burette. So, when extra NaOH is added in the flask after endpoint, the pH shoots above 8.2 giving dark pink color to the solution in flask.
In other terms, this titration is meant to standardize NaOH using KHP as the standard.
# A very dark (intense) pink color indicates that the base has been added in excess more that the actually required to attain the titration end point.
So, if dark pink color is taken as endpoint, there has been excess (more that actually required) NaOH has been added.
# Note that for p
# Suppose there were 0.020 moles of KHP in solution (flask).
Case 1: Faint Pink as end point: The actual volumes of NaOH solution (taken in burette, for standardization of NaOH) to be consumed to reach end point = 10.0 mL.
Since 1 mol NaOH neutralizes 1 mol KHP, you known that there are 0.020 moles in NaOH in the 10.0 mL NaOH released into the flask.
So,
Actual [NaOH] = 0.020 mol / (0.010 L) = 2.0 M
Case 1: Dark Pink as end point: You dispensed 12.0 mL NaOH from burette instead of 10.0 mL.
Since 1 mol NaOH neutralizes 1 mol KHP, you known that there are 0.020 moles in NaOH in the 12.0 mL NaOH released into the flask.
So,
Erroneous [NaOH] = 0.020 mol / (0.012 L) = 1.67 M
Conclusion: Therefore, if you dispensed extra (more than actual) volume of NaOH from the burette, the calculated [NaOH] would be lesser than the actual value because you added abnormally NaOH to neutralize the same number of moles of KHP.
So, correct option is – E. Lower because more volume of NaOH is being used to titrate the moles of KHP.