Molarity and Equivalents Q

[SonhosDaVida]

Why in these examples do you have to multiply the molarity by the Equivalents? And how do you find Equivalents?

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[AdaptPrep]

Hi, SonhosDaVida--

One mole of NaOH produces one mole of OH-.
One mole of H2SO4 produces two moles of H+.

We need to account for the two moles of H+ for every one mole of H2SO4. It is a way of getting the total moles of H+ available to react.

Personally, I prefer using the formula N1V1 = N2V2 where N = normality. Normality, recall, is molarity times the number of moles of H+ or OH- per mole of molecules (or, as they called it, equivalents). For example, 1.0 M H2SO4 is 2.0 N H2SO4 since one mole of H2SO4 donates 2 moles of H+. 2.0 M H3PO4 would be 6.0 N H3PO4 since one mole of H3PO4 donates 3 moles of H+. 3.0 M Ca(OH)2 is 6.0 N Ca(OH)2 since one mole of Ca(OH)2 donates 2 moles of OH-.

If you convert molarity to normality right off the bat, it makes life easier:
0.25 M H2SO4 = 0.50 N H2SO4
(100 mL)(0.50 N) = (x mL)( 0.40 N)
x = 125 mL

Here is the full stoichiometry calculation. Hopefully, you can see why we must multiply by 2.
Units will cancel and leave you with mL NaOH. Remember, molarity is moles/liter or mmoles/mL: