How many hydrogen cations will be in a solution if you start by dissolving 3.4 moles of phosphoric acid?
Phosphoric acid is a weak acid which doesn't completely ionize in solution. To answer this question you would need to know the volume of solution, as you can calculate the concentration of hydrogen ions from the acid dissociation constant of phosphoric acid if you know its concentration.
I'll explain how to calculate the number of hydrogen atoms in 3.4 moles of phosphoric acid, which is `H_3PO_4` , but recognize that even though the phosphoric acid will be dissolved in the solution the hydrogen ions won't all be separated from the phosphate ions.
(3.4 moles H3PO4)(3 H/1 H3PO4) = 10.2 moles H
10.2 moles H x (6.02 x 10^23 atoms/1 mole) = 6.14 x 10^24 H atoms
If you're looking at this on a more advanced level, involving acid equilibrium, then you need to know the Ka's for phosphoric acid:
Ka1 = 7.25 x 10^(-3)
Ka2 = 6.31 x 10^(-8)
Ka3 = 4.80 x 10^(-13)
H3PO4 has three Ka's because it's triprotic, meaning that is produces 3 hydrogen ions per molecule. The dissociations are successively weaker, meaning that each H+ comes off less easily than the one before it.
The equilibrium constant expression for the first dissociation is:
`7.25 x 10^(-3) =( [H^+])([H_2PO_4^-])/([H_3PO_4])`
Since the second Ka2 is 5 orders of magnitude smaller than the first, the second an third dissociations won't contribute significantly to the H+ concentration.
If you call the amount that dissociated "x", then at equilibrium the concentrations are:
[H+] = x, [H2PO4-] = x, [H3PO4] = initial concentration-x
and 7.25 x 10^(13) = x^2/[H3PO4-x]
You can find the initial concentration of H2PO4 by dividing 3.4 moles by the volume in liters. Solving the equation for x will give you the hydrogen ion concentration.