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 36.75 mL of the acid, and the solution turns colorless. Then, it takes 3.50 mL of the NaOH to reach the end point. (a) What is the concentration of the HNO3 solution? My answer was 1.85 X 10^-2 M. but it said that I was wrong but within 10% of the correct value Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. M (b) How many moles of NaOH were in excess after the first addition? I got 4.64 X 10^-4 and it was right but it said I needed to check my significant figures. Please help. Thank you.

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 =[ (0.02775L + 0.00398L)(0.0502M NaOH) (1mol HNO3 / 1mol NaOH) ] / (0.050 + 0.036L) M of HNO3 = 0.01852 M or 1.85 x 10^ -2 M B. moles of HNO3 = 0.01852M x 0.050L = 0.0009261 moles or 9.26 x 10^ -4 mole of NaOH added inititally = 0.02775L x 0.0502M = 0.001393 excess mole NaOH = 4.67 x 10 ^ -4 moles

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