you have 20.00mL + 14.02mL = 34.02mL of solution of pH = 4.00 after titration
That will have
.. [H+] = 10^-4.00 = 1.000x10^-4 mol H+ / L
so mole [HCl] remaining after titration would be
.. 34.02mL * (1L / 1000mL) * (1.000x10^-4 mol HCl / L) = 3.402x10^-6 mol
moles HCl reacted would be
.. 14.02 mL NaOH * (1L/1000mL) * (0.02932 mol/L) * (1mol HCl / 1 mol NaOH) = 4.1111x10^-4
so moles HCl present in 20.00mL aliquot is
.. 4.1111x10^-4 + 0.03402x10^-4 = 4.1451x10^-4
and that translates to a molarity of
.. (4.1451x10^-4 mol / 20.00mL) * (1000mL / L) = 0.020726 M HCl
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let's discuss the strong acid resins on an side note...
the strong acid resin contains SO3- groups which scavenge the Mg2+ and leave the two Cl- ions.
The way the resins work is when they are fully regenerated, they are saturated with Na+ ions. SO3- ---- Na+.
The Mg2+ ions displace the Na+ and are removed from the water stream.. SO3- ---- Mg2+ ----- SO3- Once the resins are saturated with Mg2+, they are regenerated by soaking them in brine which replaces the Mg2+ ions with Na+ and the process starts all over.
It is POSSIBLE to purchase new resins that are of the acid form and have H+ attached to the SO3- groups. H+ instead of Na+. We do it at my company. But it's atypical and not cost effective. Let's assume that's what's happening in this case so that
.. the Mg from MgCl2 gets adsorbed onto the resins and displaces H+
.. the H+ reacts with the Cl- and makes HCl which we test.
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back to the problem
the 10.00 aliquot starts with
.. 10.00mL * (1L / 1000mL) * (0.020726 mol HCl / L) = 2.0726x10^-4 mol HCl
moles of HCl remaining after titration (same deal as above.. pH = 4.00)
.. 75.00mL * (1L / 1000mL) * (1.000x10^-4 mol HCl / L) = 7.500x10^-6 mol
moles of HCl reacted with the NaOH during titration
.. 35.94mL NaOH * (1L/1000mL) * (0.02932 mol/L) * (1mol HCl / 1 mol NaOH) = 1.0538x10^-3
and we can write
.. mole HCl from acid + mol HCl from MgCl2 = mol HCl reacted + moles HCl remaining
and then
.. mol HCl from MgCl2 = (1.0538x10^-3 + 7.500x10^-6) - 2.0726x10^-4 = 8.5404x10^-4
and mol of MgCl2
.. 8.5404x10^-4 mol HCl * (1 mol MgCl2 / 2 mol HCl) = 4.2702x10^-4 mol MgCl2
and that is in a 10.00mL aliquot so that
.. (4.2702x10^-4 mol MgCl2 / 10.00mL) * (1000mL / 1L) = 4.2702x10^-2 M MgCl2
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notice I carry 1 extra sig fig during intermediate steps? that's to avoid rounding errors
FINAL answer is
.. molarity HCl = 0.02073M
.. molarity MgCl2 = 0.04270M
