Question

A 5.00 mL volume of a 0.60 M solution of phosphoric acid, H3PO4(aq), reacts with 45.00 mL of 0.25 M NaOH(aq) according to the reaction below. The temperature of the solution rises from 26.0 °C to 28.0 °C. Assume the volumes of the solutions are additive, the density of the final solution is 1.10 g mL–1 , the heat capacity of the solution is the same as water and the water produced by the reaction does not affect the volume of the resulting solution. Calculate the molar enthalpy change for the reaction.

H3PO4(aq) + 3 NaOH(aq) ® Na3PO4(aq) + 3 H2O(l) Hrxn = ?

Answer 1

Number of moles , n = Molarity x volume in L

nH3PO4 = 0.60M x 5.00 mLx10^-3 L/mL = 3x10^-3 mol

nNaOH = 0.25M x 45.00 mLx10^-3 L/mL = 0.01125 mol

Total volume of the solution is V = 5.0 mL + 45.0 mL = 50.0 mL

Density of solution , d = 1.10 g/mL

So mass of solution , m = density x volume

                                   = 1.10 g/mL x 50.0 mL

                                   = 55 g

specific heat of solution , c = 4.184 J/g-oC

change in temperature , dt = 28-26=2.0 oC

So heat liberated by the system , Q = mcdt = 460.24 J

This amount of heat was liberated by 3x10^-3 moles of H3PO4

For 1 mole of H3PO4 the heat liberated is 460.24J/(3x10^-3 mol) = 153.4x10³J = 153.4 kJ

molar enthalpy change for the reaction is -153.4 kJ