Question

You need to prepare 5.00 L of an aqueous solution having a pH = 12.50 from Ca(OH)_2. Find the [OH^-] in the solution Find the number of grams of ca(OH)_2 needed to prepare this solution (if you could't solve ,assume a k = FAKE[OH^-] = 0.0225 M. An aqueous solution is prepared by dissolving 0.125 mol of acetic acid HC_2H_2O_2 in 250.0 mL of solution. Find the % ionization of the acid at equilibrium: SHOW ALL WORK
Please help

Answer 1

The Ca(OH)₂ solution has a pH of 12.50, this means that the pOH is:
pOH = 14-12.5 = 1.5
[OH^-] = 10^-1.5 = 0.0316 M

Now, that we know this, we know that the solution is dissolved this way:
Ca(OH)₂ <--------> Ca²⁺ + 2OH^-

This means that to get the concentration of Ca(OH)₂:
[Ca(OH)₂] = 0.0316 / 2 = 0.0158 M

the moles and then mass:
moles = 0.0158 mol/L * 5 L = 0.079 moles
mass = 0.079 moles * (40+32+2) = 5.846 g needed to prepare this solution.

The other question post it in another question thread :). Hope this helps