Question

H3PO4(aq)+3KOH(aq) >> 3H2O(l) + K3PO4(aq)+ 173.2 kj

The neutralization of H3PO4 with KOH is exothermic. 70.0 mL of 0.207 M H3PO4 is mixed with 70.0 mL of 0.620 M KOH initially at 22.76 °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.

Answer 1

moles of H3PO4 = 70 x 0.207 / 1000 = 0.01449

moles of KOH = 70 x 0.620 / 1000 = 0.0434

H3PO4   + 3 KOH ----------------------> K3PO4 + 3H2O

1 mol               3 mol

0.01449          0.0434

limiting reagent is KOH

moles of water formed = 0.0434

delta H = - Q / n

-173.2 = -Q / 0.0434

Q = 7.52 kJ =7517 J

mass = 1.13 x (70 + 70) = 158.2 g

Q = m Cp dT

7517 = 158.2 x 3.78 x dT

dT = 12.6

T2 - 22.76 = 12.6

T2 = 35.33 oC

final temperature = 35.33 oC