Chem question..... titrations?

Question

"A student was titrating 15.0 mL of NaOH solution of unnknown concentration i a 100 mL beaker. After adding 17 mL of 0.01 M HCL, she realized she had forgotten to add the indicatior. She quickly added 2 drops of bromothymol blue indicatior and her solution went yellow immediately. What is the simplest procedure thi student can do to correct and finish her experimnt and not have to start over? Describe experimental procedures and mathematical analysis."

This is for my chem lab, and I'm not sure what to do. You could add more NaOH until the solution is neutralized, but that wouldn't tell you the concentration or even how much HCL you neutralized. Any ideas?

Answer 1

The yellow colour means that there is an excess of acid.
So you have a beaker with excess acid but also an indicator. This means you can do a second titration.
You have two options.

Option 1. Provided that you have lots of your NaOH solution that you wanted to titrate, fill a burrete with it and titrate what's inside the flask with this solution.
Let's assume that the unknown concentration of NaOH is C.
Then since the stoichiometry is 1:1, the amount of moles of HCl are going to be equal to the moles of NaOH that you had in the beginning plus the ones you added during the second titration. So if you added Vt ml of NaOH you have
mole HCl= mole NaOH sample +mole NaOH titration

=>C(HCl)* V(HCl)= C*V(NaOH sample)+ C*Vt
=>0.01 *17= C*15+C*Vt
=> 0.17= (15+Vt)*C
=> C=0.17/(15+Vt)

Determine Vt experimentally, substitute in the equation and solve.

Option 2. If you don't have any more NaOH of unknown concentration. Prepare a standard NaOH solution of known concentration, lets call it Cstand. It should be in the range of the HCl you added so you could use 0.01 or 0.005 M.

Then you have the same equation as in option one but now
mole NaOH titration = Cstand*Vt
so
C(HCl)* V(HCl)= C*V(NaOH sample)+ Cstand*Vt
=>0.01 *17= C*15+Cstand*Vt=>

C=(0.17-Cstand*Vt)/15

You chose Cstand, then you determine Vt experimentally and finally you can calculate C.

Note that you don't need to convert ml into liters since the conversion factor is simplified in these formulae

Answer 2

Add the amount necessary to neutralize. Do your normal concentration calculation based on the total amount of NaOH and drops of tritration. Divide the total mL of sample by 15. Divide the result of your concentrations by this amount.