Determine the hydronium ion concentration in a solution that is 0.00018 M Ca(OH)2. Answer in units of M
To determine the hydronium ion (H3O+) concentration in a solution of 0.00018 M Ca(OH)2, we need to consider the dissociation of water and the dissociation of calcium hydroxide (Ca(OH)2) in water.
The dissociation of water:
H2O ⇌ H+ + OH-
The dissociation of calcium hydroxide:
Ca(OH)2 ⇌ Ca2+ + 2OH-
Since calcium hydroxide (Ca(OH)2) dissociates to produce two hydroxide ions (OH-), we can determine the concentration of hydroxide ions by multiplying the concentration of calcium hydroxide by 2:
[OH-] = 2 * 0.00018 M = 0.00036 M
According to the principle of charge balance in water, the concentration of H3O+ ions (hydronium ions) will be equal to the concentration of hydroxide ions:
[H3O+] = [OH-] = 0.00036 M
Therefore, the hydronium ion concentration in the given solution is 0.00036 M.
Ca(OH)2 ==> Ca^2+ + 2OH^-
If it is 0.00018M in Ca(OH)2 it must be 2*0.00018 M in OH.
Then (H3O^+)(OH^-) = Kw = 1E-14. Knowing Kw and OH^- you can calculate H3O^+