Question

The neutralization of H3PO4 with KOH is exothermic. 55.0 mL of 0.213 M H3PO4 is mixed with 55.0 mL of 0.640 M KOH initially at 22.43 °C. Predict the final temperature of the solution if its density is 1.13 g/mL and its specific heat is 3.78 J/(g·°C) Assume that the total volume is the sum of the individual volumes.
RXN: H3 PO4(aq) + 3KOH(aq) 3 H2O(l) + K3PO4(aq) + 173.2 kj

Answer 1

Answer:

26.75ºC

Explanation:

By the reaction, we can notice that the reaction releases 173.2 kJ (173200 J) per mol of H₃PO₄. This heat can also be calculated by:

Heat released per mol of H₃PO₄ = msolution*csolution*ΔT/nH₃PO₄

Where msolution is the mass of the solution, csolution is the specific heat, ΔT is the variation at temperature (final - initial) and nH₃PO₄ is the number of moles of H₃PO₄.

msolution = Volume*density

msolution = (55 + 55)*1.13

msolution = 124.3 g

nH₃PO₄ = Volume * concentration

nH₃PO₄ = 0.055*0.213 = 0.011715 mol

173200 = 124.3*3.78*(T - 22.43)/0.011715

469.9674(T - 22.43) = 2029.038

T - 22.43 = 4.32

T = 26.75ºC