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.
M Ba(OH)2 solution = 0.0921 x (45.0/350.0) = ? M
Ba(OH)2 is a strong base (ionizes 100%) to concn Ba(OH)2 is ?M from above.
Then (OH^-) - twice that of the base since there are two OH ions per molecule of Ba(OH)2.
To find the concentration of OH- in the diluted solution, we need to consider the dissociation of Ba(OH)2. Ba(OH)2 dissociates into one Ba2+ ion and two OH- ions.
First, let's calculate the number of moles of Ba(OH)2 in the initial solution:
moles of Ba(OH)2 = volume × concentration
= 45.0 mL × 0.0921 M
= 4.1445 × 10^-3 moles
Since the stoichiometry of the reaction is 1:2 between Ba(OH)2 and OH-, the number of moles of OH- ions is twice the number of moles of Ba(OH)2:
moles of OH- = 2 × moles of Ba(OH)2
= 2 × (4.1445 × 10^-3 moles)
= 8.289 × 10^-3 moles
Now that we know the moles of OH-, we can calculate the concentration of OH- in the diluted solution:
concentration = moles / volume
= (8.289 × 10^-3 moles) / 350.0 mL
= 2.368 × 10^-5 M
Therefore, the concentration of OH- in the diluted solution is 2.368 × 10^-5 M.