In a titration of HNO3, you add a few drops of phenolphthalein indicator to 50.0

Question

In a titration of HNO3, you add a few drops of phenolphthalein indicator to 50.00 mL of acid in a flask. You quickly add 23.75 mL of 0.0502 M NaOH but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add 30.50 mL of the acid, and the solution turns colorless. Then, it takes 3.88 mL of the NaOH to reach the end point. What is the concentration (M) of the HNO3 solution? How many moles of NaOH were in excess after the first addition? Can you work out the steps?

Explanation / Answer

milli moles of NaOH total added = (23.75 x0.0502 ) +(3.88 x0.0502) = 1.387 milli moles of acid = (50+30.5) xM equating both we get 80.5M = 1.387 M = 0.01723 is conc of HNO3 excess NaOH moles after 1st addition = (23.75 x0.0502/1000)-(0.01723x50/1000) = 0.0003307 moles = 3.307 x10^ -4 moles

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