A conductivity titration is performed between 21.3 mL of 0.1089 M calcium hydroxide and 0.0642 M phosphoric acid. The balanced equation is below.3 Cа(ОН)2 (ag) + 2 H3РО4 (aд)—> Саз(РOa)2 (s) + 6 H20What is the theoretical volume (accepted value) of phosphoric acid (in mL) at the equivalence point? Enter your answer without units.If a conductivity of zero was reached at 23.09 mL of H3PO4 (experimental value), what is the percent error? Enter your answer without the % symbol.

The balanced reaction taking place is given as, Given: Concentration of H3PO4 = 0.0642 M The concentration of Ca(OH)2 = 0.1089 M And volume of Ca(OH)2 = 21.3 mL. From the above-balanced reaction, we can see that 3 moles of Ca(OH)2 is reacting with 2 moles of H3PO4 to reach the equivalence point. Hence the moles of H3PO4 required = Moles of Ca(OH)2 taken3 × 2Since moles = concentration × volume=> Concentration of H3PO4 × volume of H3PO4 solution required = Concentration of Ca(OH)2 × volume of Ca(OH)2 solution3 × 2Hence substituting the values we get,=> 0.0642 × volume of H3PO4 solution required = 0.1089 × 21.33 × 2 => volume of H3PO4 solution required = 24.09 mL approx. Hence the theoretical volume of phosphoric acid at the equivalence point is 24.09 mL. And the percent error = Theoretical value - Experimental valueTheoretical value × 100 = 24.09 - 23.0924.09 × 100 = 4.14 % approx.