basic chem question help?

i think you mean volumes are additive...

250 mL + 610 mL = 860 mL = 0.860 L

moles Cl = moles Cl from KCl + moles Cl from MgCl2

moles from KCl = 0.250 L x (0.640 moles KCl / L) x (1mole Cl- / 1 mole KCl) = 0.160 moles Cl-

moles from MgCl2 = 0.610 L x (0.395 moles MgCl2 / L) x (2 moles Cl- / 1 mole MgCl2) = 0.482 moles Cl-

total moles Cl- = 0.160 + 0.482 = 0.642 moles Cl-

finally molarity Cl- = 0.642 moles Cl- / 0.860L = 0.746 M Cl-

**** also ****

idea here is moles H+ = moles OH- at endpoint

0.04655 L x (5.22x10^-2 moles HNO3 / L) x (1 mole H+ / 1 mole HNO3) = 2.430 x10^-3 moles H+

so we need 2.430 x10^-3 moles OH-

2.430 x10^-3 moles OH- x (1 mole Ba(OH)2 / 2 moles OH-) x (1 L / 8.52x10^-2 moles Ba(OH)2) x (1000 mL / L) = 14.3 mL

I assume that this problem is asking for the chloride concentration. So - calculate the total number of moles of chlorine and divide by the total volume.

(0.250 L)(0.640 mol KCl/L)(1 mol Cl/mol KCl) = 0.160 mol Cl
(0.610 L)(0.395 mol MgCl2/L)(2 mol Cl/mol MgCl2) = 0.482 mol Cl

So that is

0.160 mol Cl + 0.482 mol Cl = 0.642 mol Cl

total. Now divide by the total volume:

(0.642 mol Cl)/(0.250 L + 0.610 L) = 0.747 mol Cl/L

So to three significant digits, it is 0.747 M in Cl-.