Question #28634

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?

Expert's answer

3Сa(OH)_{2} + 2H_{3}PO_{4} = Ca_{3}(PO_{4})_{2} + 6H_{2}O

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 H_{3}PO_{4} solution is required to completely neutralize 22.5 ml of the Ca(OH)_{2}.

The amount of H_{3}PO_{4} could be calculated as:

v = C(M) x V

v(H_{3}PO_{4}) = 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(H_{3}PO_{4})/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

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 H

The amount of H

v = C(M) x V

v(H

According to the equation the total amount of Сa(OH)

v(Сa(OH)

So the concentration of Сa(OH)

C(Сa(OH)

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