Question

The neutralization of H3PO4 with KOH is exothermic. 70.0 mL of 0.227 M H3PO4 is mixed with 70.0 mL of 0.680 M KOH initially at 23.41 °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.

H₃PO₄(aq)+3KOH(aq) ---> 3H₂)(l)+K₃PO₄(aq)+173.2 kJ

Answer 1

delta H = - 173.2 kJ

moles of H3PO4 = 70 x 0.227 / 1000 = 0.01589

moles of KOH = 70 x 0.680 / 1000 = 0.0476

H₃PO₄(aq)+3KOH(aq) ---> 3H₂)(l)+K₃PO₄(aq)

    1                3

0.01587      0.0476

Q = - n x delta H = - 0.01587 x - 173.2

    = 2.75 kJ

    = 2749 J

volume of solution = 70 +70 = 140mL

mass = 140 x 1.13 = 158.2 g

Q = m Cp dT

2749 = 158.2 x 3.78 x dT

dT = 4.60

T2 - T1 = 23.41

T2 - 4.60 = 23.41

T2 = 28.0 oC

final temperature = 28.0 oC