If 488 mL of 2.9 molar HCl solution is added to
191 mL of 2.8 molar Ba(OH)2 solution, what
will be the molarity of BaCl2 in the resulting
solution?
Answer in units of M.
I answered a question quite similar to this a couple of days ago. There is no BaCl2 in solution. There are Ba ions and chloride ions and there are more chloride ions than barium ions. But to try to calculate a M of an ionic compound in which the positive barium ions and the negative chloride ions are not equal is nonsense.
To find the molarity of BaCl2 in the resulting solution, we need to use the concept of stoichiometry and the balanced chemical equation between HCl and Ba(OH)2.
The balanced chemical equation is:
2HCl + Ba(OH)2 -> BaCl2 + 2H2O
We have the volume and molarity of the HCl solution (488 mL and 2.9 M, respectively) and the volume and molarity of the Ba(OH)2 solution (191 mL and 2.8 M, respectively).
Step 1: Calculate the moles of HCl and Ba(OH)2 using the given volumes and molarities.
Moles of HCl = Volume of HCl solution (L) x Molarity of HCl solution (mol/L)
Moles of HCl = 0.488 L x 2.9 mol/L = 1.4152 mol
Moles of Ba(OH)2 = Volume of Ba(OH)2 solution (L) x Molarity of Ba(OH)2 solution (mol/L)
Moles of Ba(OH)2 = 0.191 L x 2.8 mol/L = 0.5348 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, compare the moles of HCl and Ba(OH)2. The reactant with the lesser number of moles is the limiting reactant.
In the balanced equation, we see that 2 moles of HCl react with 1 mole of Ba(OH)2.
So, 2 moles of HCl would react with 1 mole of Ba(OH)2.
Since we have 1.4152 moles of HCl and only 0.5348 moles of Ba(OH)2, Ba(OH)2 is the limiting reactant.
Step 3: Calculate the moles of BaCl2 formed (according to stoichiometry).
From the balanced equation, we see that 1 mole of Ba(OH)2 produces 1 mole of BaCl2.
Therefore, 0.5348 moles of Ba(OH)2 would produce 0.5348 moles of BaCl2.
Step 4: Calculate the volume of the resulting solution.
The total volume of the resulting solution is the sum of the volumes of the HCl and Ba(OH)2 solutions.
Total volume = Volume of HCl solution + Volume of Ba(OH)2 solution
Total volume = 0.488 L + 0.191 L = 0.679 L
Step 5: Calculate the molarity of BaCl2 in the resulting solution.
Molarity of BaCl2 (M) = Moles of BaCl2 / Volume of resulting solution (L)
Molarity of BaCl2 = 0.5348 mol / 0.679 L = 0.787 M
Therefore, the molarity of BaCl2 in the resulting solution is 0.787 M.