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

Number of moles of H₃PO₄ is , n = Molarity x volume in L

                                                 = 0.213 M x 0.055 L

                                                 = 0.0117 mol

Number of moles of KOH is , n' = Molarity x volume in L

                                              = 0.640 M x 0.055 L

                                              = 0.0352 moles

                                              = 3x0.0117 moles

                                              = 3 x number of moles of H₃PO₄

The balanced reaction is : H₃ PO₄(aq) + 3KOH(aq) 3 H₂O(l) + K₃PO4(aq) + 173.2 kJ

According to the balanced equation ,

1 mole of H₃PO₄ upon reaction produces 173.2 kJ of heat

0.0117 moles of H₃PO₄ upon reaction produces M kJ of heat

M = ( 0.0117x173.2) / 1

    = 2.03 kJ

This amount of heat is utilized to heat solution.

Total volume of the solution , V = 55.0+55.0 = 110 mL

Given density of the solution , d = 1.13 g/mL

So mass of the solution , m = volume x density

   = 110 mL x 1.13 (g/mL)

   = 124.3 g

So heat absorbed by solution , Q = mcdt

Where

m = mass of the solution = 124.3 g

c = specific heat of solution = 3.78 J/(g·°C)        

dt = change in temperature = final - initial

= t - 22.43 oC

Q = heat absorbed = 2.03 kJ = 2.03x1000 J = 2030 J

Plug the values we get

2030 = 124.3 x 3.78 x (t-22.43)

t = 26.75 ^oC

Therefore the final temperature of the solution is 26.75 ^oC