If 435 mL of 2.7 molar HCl solution is added to 128 mL of 3.2 molar Ba(OH)2
solution, what will be the molarity of BaCl2 in the resulting solution?
Answer in units of M
Do this like the NaCl M problem. Post your work if you get stuck.
To find the molarity of BaCl2 in the resulting solution, we need to determine the number of moles of BaCl2 formed.
First, let's calculate the number of moles of HCl and Ba(OH)2 in their respective solutions.
Number of moles of HCl = volume of HCl solution (in liters) × molarity of HCl solution
= 435 mL ÷ 1000 mL/L × 2.7 mol/L
= 1.1755 moles of HCl
Number of moles of Ba(OH)2 = volume of Ba(OH)2 solution (in liters) × molarity of Ba(OH)2 solution
= 128 mL ÷ 1000 mL/L × 3.2 mol/L
= 0.4096 moles of Ba(OH)2
From the balanced chemical equation, we know that 2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of BaCl2 and 2 moles of H2O.
Since the ratio of HCl to Ba(OH)2 is 2:1, the limiting reactant is Ba(OH)2. This means that all of the Ba(OH)2 will react and we will have 0.4096 moles of BaCl2 in the resulting solution.
Next, we calculate the total volume of the resulting solution by adding the volumes of the two solutions:
Total volume of resulting solution = volume of HCl solution + volume of Ba(OH)2 solution
= 435 mL + 128 mL
= 563 mL
Finally, we can find the molarity of BaCl2 by dividing the number of moles of BaCl2 by the total volume of the resulting solution (in liters):
Molarity of BaCl2 = number of moles of BaCl2 ÷ total volume of resulting solution (in liters)
= 0.4096 moles ÷ 563 mL ÷ 1000 mL/L
≈ 0.727 M
Therefore, the molarity of BaCl2 in the resulting solution is approximately 0.727 M.