Final answer:
To completely neutralize the Ba(OH)2 solution, we need to calculate the amount of HCl needed. The correct answer is Option 1: 26.5 mL of 0.100 M HCl.
Step-by-step explanation:
To completely neutralize the Ba(OH)2 solution, we need to calculate the amount of HCl needed. From the given information, we know that the initial concentration of HCl is 0.100 M and the initial volume is not provided. However, we can calculate the initial moles of HCl using the initial concentration and volume. Then, using the balanced chemical equation between HCl and Ba(OH)2, we can determine the moles of Ba(OH)2 present in the given volume. Finally, we can use the molar ratio to calculate the moles of HCl needed to neutralize the Ba(OH)2 solution.
Let's start by calculating the initial moles of HCl:
Initial moles of HCl = initial concentration x initial volume = 0.100 M x x mL = x mol
Next, we can use the balanced chemical equation:
2HCl + Ba(OH)2 → BaCl2 + 2H2O
From the equation, we can see that 2 moles of HCl are required to neutralize 1 mole of Ba(OH)2. So, the moles of Ba(OH)2 present in 53.0 mL can be calculated as:
Moles of Ba(OH)2 = (x mol HCl) / 2 = (x mol HCl) / 2 = (x mol HCl) / 2
Given that the concentration of Ba(OH)2 is 0.107 M and the volume is 53.0 mL, the moles of Ba(OH)2 can be calculated as:
Moles of Ba(OH)2 = concentration x volume = 0.107 M x 53.0 mL = 0.107 moles
Now, using the molar ratio, we can calculate the moles of HCl needed:
Moles of HCl = 2 x moles of Ba(OH)2 = 2 x 0.107 moles = 0.214 moles
To convert the moles of HCl to milliliters, we can use the final concentration of HCl, which is 0.100 M:
Final volume of HCl = moles of HCl / final concentration = 0.214 moles / 0.100 M = 2.14 mL
Therefore, the correct answer is Option 1: 26.5 mL of 0.100 M HCl is needed to completely neutralize 53.0 mL of 0.107 M Ba(OH)2 solution.