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 26.50 mL of 0.0502 M NaOH but overshoot the end point, and the solution turns deep pink. Instead of starting over, you add 31.25 mL of the acid, and the solution turns colorless. Then, it takes 3.28 mL of the NaOH to reach the end point. (a) What is the concentration of the HNO3 solution? Your response differs from the correct answer by more than 10%. Double check your calculations. M (b) How many moles of NaOH were in excess after the first addition?

Explanation / Answer

the chemical reaction for your titration is HNO3 + NaOH = H2O + NaNO3 M = mole / vol in L mole = M x vol in L . A. M of HNO3 =[ (0.0265L + 0.00328L)(0.0502M NaOH) (1mol HNO3 / 1mol NaOH) ] / (0.050 + 0.03125L) M of HNO3 = 0.01840 M or 1.84 x 10^ -2 M B. moles of HNO3 = 0.0184 M x 0.050L = 0.000921 moles or 9.2 x 10^ -4 mole of NaOH added inititally = 0.0265L x 0.0502M = 0.0013303 excess mole NaOH = 4.1 x 10 ^ -4 moles

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