Find the concentration of OH- of a solution of 45.0mL of 0.0921 M Ba(OH)2 diluted with enough water to make 350.0 mL of solution.
We can first find the moles of Ba(OH)2 in the 45.0 mL solution, and then divide by the total volume of the final solution (350.0 mL) to find the new concentration.
Moles of Ba(OH)2 = (0.0921 mol/L) * (45.0 mL) * (1 L/1000 mL)
Moles of Ba(OH)2 = 0.004145 moles
Now we need to account for the fact that Ba(OH)2 releases 2 moles of OH- ions for every mole of Ba(OH)2. So, the moles of OH- will be 2 * (moles of Ba(OH)2).
Moles of OH- = 2 * 0.004145 moles
Moles of OH- = 0.00829 moles
Next, we find the concentration of OH- in the final solution:
Concentration of OH- = moles of OH- / volume of final solution
Concentration of OH- = 0.00829 moles / 350.0 mL * (1 L/1000 mL)
Concentration of OH- = 0.0237 M
To find the concentration of OH- in the resulting solution, we need to consider the stoichiometry of the reaction between Ba(OH)2 and water.
The balanced equation for this reaction is:
Ba(OH)2(aq) ⇌ Ba2+(aq) + 2OH-(aq)
From the equation, we can see that each mole of Ba(OH)2 produces two moles of OH-. Therefore, the number of moles of OH- in the original 45.0 mL of 0.0921 M Ba(OH)2 solution can be calculated as follows:
moles of OH- = 0.0921 M * 0.0450 L = 0.0041455 mol
Since the solution is then diluted to a total volume of 350.0 mL, the final concentration of OH- can be calculated as follows:
final concentration of OH- = moles of OH- / total volume of the solution
final concentration of OH- = 0.0041455 mol / 0.350 L = 0.0118 M
Therefore, the concentration of OH- in the resulting solution is 0.0118 M
To find the concentration of OH- in the solution, we need to first understand the stoichiometry of the reaction between Ba(OH)2 and water. Ba(OH)2 dissociates to form Ba2+ ions and two OH- ions.
The balanced equation for the dissociation of Ba(OH)2 is:
Ba(OH)2 → Ba2+ + 2OH-
From this equation, we can determine that every mole of Ba(OH)2 yields two moles of OH-.
Step 1: Calculate the moles of Ba(OH)2:
Moles of Ba(OH)2 = concentration × volume
Moles of Ba(OH)2 = 0.0921 M × (45.0 mL / 1000 mL)
Moles of Ba(OH)2 = 0.0041445 mol
Step 2: Calculate the moles of OH-:
Moles of OH- = 2 × Moles of Ba(OH)2
Moles of OH- = 2 × 0.0041445 mol
Moles of OH- = 0.008289 mol
Step 3: Calculate the concentration of OH-:
Concentration of OH- = Moles of OH- / Volume
Concentration of OH- = 0.008289 mol / (350.0 mL / 1000 mL)
Concentration of OH- = 0.0237 M
Therefore, the concentration of OH- in the solution is 0.0237 M.