Question

So I did a titration lab where were reacted various amounts of Mg with 10mL of HCL and then neutralized the raction with NaOH. I have my table of values for Mass of Mg (g), Initial Volume NaOH (mL), Final Volume NaOH (mL), End Point Volume (mL), Limiting Reagent (Mg or HCL). There is a Titration table I need to fill in that asks for Moles of Mg (mol), Moles of NaOH (mol) and Moles of H⁺ reacting with Mg (mol). I have the Moles of Mg calculated and I also have Moles of NaOH (mol) calculated but how do I go about finding Moles of H⁺ reacting with Mg. The manual says I can use: Moles of H⁺ total - moles of H⁺ excess = moles H⁺ reacting with metal. Where could I obtain those numbers.

I had in my notes to take Moles of Mg - Moles of NaOH and that gives the value but the problem with I am getting with that method is that if my limiting reagent is Mg then my Moles of H⁺ calculation ends up being a negative number.

My formula for calculating Moles of Mg is (Mass of Mg starting) * (1Mol/24.31g) = Moles of Mg

My formula for Calculating Moles of NaOh is: (End Point Volume)*(1L/1000mL)*(.3M HCL/1L) = Moles of NaOH

Thanks!

Some examples of my numbers would be

Moles of Mg = .001, Moles of NaOH = 8.58*10^-3  so that gives me a -7.6*10^-3 but I don't know if that is correct. Any help?

Answer 1

If you know the number of moles of H+ unreacted, then you can calculate the number of moles of H+ that are reacted as you stated,

moles of H+ reacted = (total moles of H+) - ( moles of H+ unreacted )

And another way to calculate the number of moles of H+ is,

Mg(s) + 2HCl ----------> MgCl2 + H2(g)

This is the reaction between your Mg and HCl.

From the balanced equation, no.of moles of Mg = 1/2 moles of H+ reacting

=> moles of H+ reacting = 2 x moles of Mg