Please explain the process: A 35.0 mL sample of 0.846 M Ba(OH)2 is added to 25.0 mL of 0.150 M HCl. What are the concentrations of the excess ions and the spectators when the solutions are allowed to react? (Assume the volumes are additive).

Sigh! Always me doing these!!! This will be rather long, so pay attention. Balanced equation.

Ba(OH)₂ + 2HCl --> BaCl₂ + H₂O

Need moles both reactant species. Molarity = moles of solute/Liters of solution. ( 35.0 mL = 0.035 Liters and 25.0 mL = 0.025 Liters )

0.846 M Ba(OH)₂ = X moles/0.035 L

= 0.02961 moles Ba(OH)₂ ------

0.150 M HCl = X moles/0.025 L

= 0.00375 moles HCl ------

Now, find limiting reactant. ( I hope HCl )

0.00375 moles HCl (1 mole Ba(Oh)₂/2 moles HCl)

= 0.001875 moles Ba(OH)₂ ------So, you have much more barium hydroxide than this. (0.02961 moles. This means HCl limits and will drive reaction )( Check the other way and see that this is true )

0.00375 moles HCl (1 mole BaCl₂/2 mole HCl)(208,2 grams/1 mole BaCl₂)

= 0.3904 grams BaCl₂ ------Now, get moles BaCl₂
0.3904 grams BaCl2 (1 mole BaCl2/208.2 grams)

= 0.001875 moles BaCl₂ ------( Now, 0.035 liters and 0.025 L = 0.06 Liters )

Molarity =  0.001875 moles BaCl₂/0.06 Liters)

= 0.031 M BaCl₂ ====== (answer in part )

This is the molarity of the barium and you can figure out the molarity of the chlorine yourself. The rest is just water and is just adding solution to solution, though it does not count in our calculations.