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
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.
1 answer
1 answer