What is the concentration of a Ca(OH)2 solution if 15.0 ml of 0.500 M H3PO4 solution is required to completely neutralize 22.5 ml of the Ca(OH)2 solution?
3Сa(OH)2 + 2H3PO4 = Ca3(PO4)2 + 6H2O C(M) = v/V C(M)-molar concentration of the solution; v - molar mass, mole; V - volume of the solution. v = m/M m - mass of the substance, grams; M - molar mass of the substance.
From the task we know that 15.0 ml of 0.500 M H3PO4 solution is required to completely neutralize 22.5 ml of the Ca(OH)2. The amount of H3PO4 could be calculated as: v = C(M) x V v(H3PO4) = 0.5 x 0.015 = 0.0075 mol According to the equation the total amount of Сa(OH)2 is: v(Сa(OH)2) = v(H3PO4)/2*3 = 0.0075/2*3 = 0.011 mol
So the concentration of Сa(OH)2 is: C(Сa(OH)2) = v/V = 0.011/0.0225 = 0.48 M