Final answer:
The molarity of the original Ba(OH)2 solution was found by reverse dilution calculations, which showed that after considering the contributions of NaOH, the Ba(OH)2 provided 0.0010 moles in 100.00 mL, resulting in a molarity of 0.0100 M.
Step-by-step explanation:
To calculate the molarity of the original Ba(OH)2 solution, we need to work through several dilution steps backwards. Firstly, the concentration of OH- ions in Solution B is given as 2.00 x 10-3 mol/L.
Since Solution B was made by diluting 25.00 mL of Solution A to 250.00 mL, the concentration of OH- ions in Solution A was 10 times greater, that is, 2.00 x 10-2 mol/L.
Solution A was produced by mixing 100.00 mL of Ba(OH)2 with 500.00 mL of 0.0200 M NaOH. The total volume of Solution A is 600.00 mL.
Assuming all the OH- ions in Solution A come from NaOH and Ba(OH)2,
we can use the formula Concentration = Moles / Volume to find the moles of OH- from NaOH: 0.0200 M NaOH * 0.5000 L = 0.0100 moles NaOH.
As NaOH provides one OH- ion per molecule, moles of NaOH equals moles of OH- ions from NaOH.
Since Solution A contains 2.00 x 10-2 mol/L of OH- ions and the volume is 0.6000 L, the total moles of OH- in Solution A is 0.0120 moles.
Subtracting the moles contributed by the NaOH (0.0100 moles), we find that 0.0020 moles of OH- ions are provided by Ba(OH)2.
Because Ba(OH)2 dissociates to provide 2 OH- per molecule, the moles of Ba(OH)2 are 0.0010 moles
(half of 0.0020 moles).
To find the molarity of the original Ba(OH)2 solution,
we divide these moles by the original volume of the Ba(OH)2 solution: 0.0010 moles / 0.1000 L,
resulting in a molarity of 0.0100 M Ba(OH)2.