Balance the chemical equation given below, and determine the number of millilitres of 0.00300 mol L^-1 phosphoric acid required to neutralize 45.00 mL of 0.00150 mol L^-1 calcium hydroxide.
________ Ca(OH) ₂(aq) + ________ H₃PO₄(aq) → ________ Ca₃(PO₄) ₂(aq) + ________ H₂O(l)

Question

Quizplus · Accepted Answer

The balanced chemical equation is:

Ca(OH)2(aq) + 2H3PO4(aq) → Ca(H2PO4)2(aq) + 2H2O(l)

From the equation, we can see that one mole of Ca(OH)2 reacts with two moles of H3PO4.

First, we need to calculate the number of moles of Ca(OH)2:

Moles of Ca(OH)2 = volume x concentration
Moles of Ca(OH)2 = 0.045 L x 0.00150 mol/L
Moles of Ca(OH)2 = 6.75 x 10^-5 mol

Since two moles of H3PO4 are required to react with one mole of Ca(OH)2, we need:

Moles of H3PO4 = 2 x moles of Ca(OH)2
Moles of H3PO4 = 2 x 6.75 x 10^-5 mol
Moles of H3PO4 = 1.35 x 10^-4 mol

Finally, we can calculate the volume of 0.00300 mol/L H3PO4 needed:

Volume of H3PO4 = moles of H3PO4 / concentration
Volume of H3PO4 = 1.35 x 10^-4 mol / 0.00300 mol/L
Volume of H3PO4 = 0.045 L or 15.0 mL

Therefore, the answer is B) 15.0 mL.

Balance the Chemical Equation Given Below, and Determine the Number