Calculate the molar heat of neutralization in kj/mol of the reaction between HA and BOH given the following information:

Question

The temperature change equals 9C
50mL of 1M concentration of Acid
50ml of 1M concentration of Base
Heat Capacity of the calorimeter is 6.5 J/C.
Specific heat is 4.18 J/gC

I can work the problem and get the correct answer of 76.41 but during the steps to get the answer, how do I know to divide by 0.5?

Answer 1

I don't know how you worked it but I would do this.
50 mL x 1M + 50 mL x 1M = 100 mL x 0.5 M (the 1M solution is diluted for both solutions) = 0.1 L x 0.5 M = 0.05 moles.
Total q = (100 x 4.18 x 9) + (6.5 x 9) = 3820.5 J.
Then 3820.5 J/0.05 moles = 76,410.5 J/mol or 76.41 kJ/mol