Question

Our chemist found a solution of 0.333M H3PO4 (phosphoric acid) and another of 1.05M NaOH (amazing what you can find in the belly of a ship) and so she did an enthalpy of neutralization reaction using 50.0mL of each solution. (Phosphoric acid is a triprotic acid; each mole of acid can yield 3 moles of H* ions.) Using the CRC Handbook at hand, the density and specific heats of the final solution were found to be 1.11 g/mL and 3.81 J/ g °C, respectively. From the graph of her results, she determined the average temperatures of the acid and basic solutions prior to mixing, an initial temperature of 20.1 °C. In the same manner, she determined that the neutralized solution, her product, reached a maximum temperature of 26.0 °C at the time of mixing. Using the calorimeter constant, calculate the magnitude (i.e. absolute value only) of the heat lost to the environment, AHeny in Joules, for the reaction of phosphoric acid and sodium hydroxide. AHen Calorimeter constant x ATrxn Enter your answer: Calculate the AHneut in Joules using the equation: AHneut AHn AHenv Enter your answer:

Answer 1

50 ml of 0.333M H₃PO₄ will react with 50 ml of 1 M NaOH (since 1 mole of H₃PO₄ consumes 3 moles of NaOH)

1/3 H₃PO₄ + NaOH ----> 1/3 Na₃PO₄ + H2O   $\Delta$H = -55.84 kJ

Total volume of solution = 100 ml

density of solution = 1.11 g/cc

specific gravity = 3.81 J/g ^oC

Initial temperature = 20.1^o C

Final Temperature = 26.0^o C

Heat of the solution = mc$\Delta$T

= 1.11 * 100 * 3.81 * (26-20.1)

= 2.495 kJ

Heat of the reaction = - heat of the solution = -2.495 kJ

Enthalpy of neutralization = - 2.495 * 1000/50 (since we used only 50 ml of 1M NaOH instead of 1000 ml)

Enthalpy of neutralization = -49.9 kJ/mol