To find the molarity of the Ba(OH)2 solution, we can use the equation:
Molarity1 x Volume1 = Molarity2 x Volume2
Where Molarity1 and Volume1 represent the initial concentration and volume, and Molarity2 and Volume2 represent the final concentration and volume.
Given:
Volume1 of Ba(OH)2 solution = 25 mL
Volume2 of HCl solution = 35 mL
Molarity2 of HCl solution = 0.1 M
We can assume that the reaction between Ba(OH)2 and HCl is a 1:2 ratio based on the balanced chemical equation:
Ba(OH)2 + 2HCl -> BaCl2 + 2H2O
Since 1 mole of Ba(OH)2 reacts with 2 moles of HCl, the moles of HCl used can be calculated as follows:
Moles of HCl = Molarity2 x Volume2
Moles of HCl = 0.1 M x 0.035 L (since 35 mL = 0.035 L)
Moles of HCl = 0.0035 mol
Since the reaction is a 1:2 ratio, the moles of Ba(OH)2 used would be half the moles of HCl:
Moles of Ba(OH)2 = 0.0035 mol / 2
Moles of Ba(OH)2 = 0.00175 mol
Now, we can calculate the molarity of the Ba(OH)2 solution:
Molarity1 = Moles of Ba(OH)2 / Volume1
Molarity1 = 0.00175 mol / 0.025 L (since 25 mL = 0.025 L)
Molarity1 = 0.07 M
Therefore, the molarity of the Ba(OH)2 solution is 0.07 M.
The answer option that matches this result is (C) 0.07.
I hope this explanation clarifies the calculation. Let me know if you have any further questions!