Chemistry
posted by Sally on .
The neutralization of H3PO4 with KOH is exothermic.
H3PO4(aq) + 3KOH(aq) > 3H2O(l) +K3PO4(aq) +173.2KJ
70.0 mL of 0.207 M H3PO4 is mixed with 70.0 mL of 0.620 M KOH initially at 21.87 °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.

You have 70.0mL x 0.207 M = 14.49 millimols which will use 43.47 mmols KOH.
You have 70.0mL x 0.620M = 43.4 mmols KOH. So the reaction will produce 43.4 mmols (0.0434 mols product) or
173,200 J/mol x 0.0434 = ? J.
?J = mass soln x specific heat soln x (TfinalTinitial)
mass soln = (70mL+70mL)*1.13 g/mL
Solve for Tfinal. 
I got 7516.88J for ?J
and mass of soln = 158.2g
So my equation is
7516.88 J = 158.2g x 3.78 J/(g·°C)x(Tfinal 21.87)
12.57 = Tfinal  21.87
Tfinal = 34.44°C
Which was wrong.
The hint they gave me was
Start by finding the amount of heat produced when 70.0 mL of 0.207 M H3PO4 react with 70.0 mL of 0.620 M KOH. Use the total volume and the density to find the mass. Then use the mass, m, heat, q, and specific heat, c, to find the change in temperature, ΔT.
Any idea as to what went wrong?