You are titrating a 50.0 mL sample of HNO3 with 0.20 M NaOH.

Question

After adding 45.3 mL of the base you remember that you have not added phenolphthalein indicator. Upon adding the indicator, the solution turns bright pink. You decide to “back-titrate” by adding 0.10 M HCl to the solution. You reach the stoichiometric point by adding 11.6 mL of the 0.10 M HCl. Determine the concentration of the original HNO3 solution.

Answer 1

First convert the NaOH sample you added to Mols

45.3mL/1000mL = .0453L*.2M = .00906mol NaOH

giving us .00906mol OH-

then convert the HCl You added

11.6mL/1000mL = .0116L*.1M = .00116mol HCl

giving us .00116mol H+

to neutralize the HCl we added we would then need .00116mol OH-

(H+ + OH-  --> H2O)

so .00906mol OH- - .00116mol OH- = .00790mol OH-

In order for our solution to be neutral we don't want to have any OH- "remaining" so the what wasn't neutralized by the HCl must have been neutralized by the HNO3 we can assume because Nitric Acid is mono-protic (Has one Hydrogen) that the remaining mol of OH- is equal to the mols of HNO3 (1 H+ is needed per OH-, if the acid was H2SO4 we would divide the mol OH by 2 to get to mols of H2SO4) so we have

.00790mol H+ from the Nitric Acid. to get concentration  convert the 50ml to L then divide the mols you have by the volume you just calculated

50.0mL/1000mL = .0500L

.00790mol H+ / 0.0500L = .158M HNO3

Answer 2

45.3 mL x 0.20 M = 9.06 mmol

11.6 mL x 0.10 M = 1.16 mmol

net NaOH = 9.06 - 1.16 = 7.90 mmole

50.0 mL HNO3 x Molarity = 7.90 mmole

HNO3 = 0.158 M