Ex e: A 25.00 mL of HCI solution was titrated to the phenolphthalein end point.

Question

Ex e: A 25.00 mL of HCI solution was titrated to the phenolphthalein end point. If 23.85 mL of 0.099 M NaOH was use d, what was the molarity of HCl solution? Calculate the number of grams of HCI in 100.0 mL of concentrated HCI which is ap 12.1 M and 37 % HCl by weight and the density of 1.191 g/mL. 2. proximately An acid-base titration is performed but some base that splashed onto the sides of the flask is not rinsed down prior to the endpoint measurement. Will the calculated molarity of the acid sample be erroneously high or low? Explain. 4. An acid solution is titrated to the pink phenolphthalein end point by a base solution. As time goes by, the pink color disappears and the solution becomes colorless again. This takes about 30 seconds or so and definitely is not due to the solutions mixing too slowly. Explain why this occurs. What is the relevance of acid-base titration? Write a few paragraphs discussing the rele acid base titration to one of the following examples 5.

Explanation / Answer

1)

volume of HCl = 25.00 mL

volume of NaoH = 23.85 mL

Molarity of NaOH = 0.0991 M

millimoles of NaOH = 0.0991 x 23.85 = 2.36

NaOH   +    HCl    ------------> NaCl   + H2O

moles of NaOH = moles of HCl

2.36 = 25 x M

Molarity of HCl = 0.0945 M

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