[Tybalt]

I was running through my numbers to complete my TITRATION lab report and realized that I messed up right out of the gate.  In creating a 0.1 M CuCl₂ * 2H₂O solution  I used 1.525g instead of 1.705g, shorting my solution about 0.180g. Making the concentration less than 0.1M, right?

My main question, is there some way to mathematically calculate the correct amount (mL) of NaOH I was to burette into the CuCl₂ * 2H₂O solution?

Redoing the lab isn't a possibility...

[Guitarmaniac86]

I was running through my numbers to complete my TITRATION lab report and realized that I messed up right out of the gate.  In creating a 0.1 M CuCl₂ * 2H₂O solution  I used 1.525g instead of 1.705g, shorting my solution about 0.180g. Making the concentration less than 0.1M, right?

My main question, is there some way to mathematically calculate the correct amount (mL) of NaOH I was to burette into the CuCl₂ * 2H₂O solution?

Redoing the lab isn't a possibility...

Yes there is.

I can give you the formula to manipulate but then you wont really learn anything other than manipulating results.

I got in serious trouble with my university for mathematically working out the correct values for a titration.

Edit:

Use the results you have and in the evaluation, explain what went wrong.

Edit 2:

I feel bad now.

Heres a hint: How do you work out concentrations, volumes and moles from a titration? If you know that, you have your answer. Also, look at the reaction... It will help if its balanced and something is worked out to do with ratios of something....

[Tybalt]

To some extent your right, it wouldn't be very constructive if you had handed me the information before I preformed my experiment, to which I would have blown off the lab completely.  But given that I've already gone through the experiment and learned the error of my mistake, theres no harm in helping correct my error.

[Tybalt]

Use the results you have and in the evaluation, explain what went wrong.

I didn't even think of that, I think I'll use it.

[Guitarmaniac86]

To some extent your right, it wouldn't be very constructive if you had handed me the information before I preformed my experiment, to which I would have blown off the lab completely.  But given that I've already gone through the experiment and learned the error of my mistake, theres no harm in helping correct my error.

Well...

OK.

CuCl₂ . 2H₂O + 2NaOH --> Cu(OH)₂ + 2NaCl + 2H₂O

The reaction is 1 : 2

Use Moles = Concentration x Volume (cm³) / 1000

To work out the number of moles of CuCl₂ . 2H₂O

However many moles of that you have, you should have 2 x number of moles for NaOH

Subsitute that into the equation:

Volume = Moles x 1000 / Concentration (volume in cm³

This should give you the theoretical volume for the NaOH

You havent given the volume so I cant give numerical values.

Sorry if I came across arsey earlier. I meant nothing by it.

[Tybalt]

No harm no foul.  I owe you one.


-Tybalt