You have 10mL of a 1.0 M Ba(OH)2 solution and add it to 2.5L of water. What is the concentration of hydroxide ions in the diluted Ba(OH)2 solution?
To find the concentration of hydroxide ions in the diluted Ba(OH)2 solution, we need to consider the dilution that occurs when adding water to the original solution.
First, let's determine the moles of Ba(OH)2 in the original 10 mL solution.
Given:
- Volume of original solution (V1) = 10 mL
- Concentration of Ba(OH)2 in the original solution (C1) = 1.0 M
Using the formula:
moles of solute (n) = concentration (C) × volume (V)
n1 = C1 × V1
= 1.0 M × 10 mL
= 0.01 moles Ba(OH)2
Next, we need to determine the final volume of the solution after dilution.
Given:
- Volume of water added (V2) = 2.5 L
The final volume (Vf) will be the sum of the original solution volume (V1) and the volume of water added (V2). However, since the volume is in different units (mL and L), we need to convert V1 to liters:
V1 (in L) = 10 mL × (1 L / 1000 mL) = 0.01 L
Vf = V1 + V2
= 0.01 L + 2.5 L
= 2.51 L
Now we can calculate the final concentration of Ba(OH)2 in the diluted solution:
Cf = n1 / Vf
= 0.01 moles / 2.51 L
≈ 0.004 L
Since Ba(OH)2 dissociates into two hydroxide ions (OH-) per molecule, the concentration of hydroxide ions in the diluted Ba(OH)2 solution is twice the concentration of Ba(OH)2:
Concentration of OH- ions = 2 × Cf
= 2 × 0.004 M
= 0.008 M
Therefore, the concentration of hydroxide ions in the diluted Ba(OH)2 solution is approximately 0.008 M.