Question

If two solutions both initially at room temperature (25 degrees Celsius): Solution 1: 35.00 mL of 0.750M HCl, Solution 2: 55.0 mL solution of 0.550 M NaOH are mixed together, what is the temperature of the resulting solution? (All usual assumptions apply)

Answer 1

The heat of neutralization of strong acid and strong base is -57.62 kJ/mol

Number of moles , n = Molarity x volume in L

nHCl = 0.750 M x 35.00mLx10-3L/mL = 0.02625 moles

nNaOH = 0.550M x 55.0 mLx10-3L/mL = 0.03025 moles

HCl + NaOH ---> NaCl + H2O

1 mole of HCl reacts with 1 mole of NaOH

0.02625 moles of HCl reacts with 0.02625 moles of NaOH

Total volume of the solution , V = 35.00+55.00 = 90.00 mL

Density of the solution , d = 1.0 g/mL

So mass of the solution , m = V x d = 90.000 mLx 1.0 g/mL = 90.0 g

The amount of heat leberated by 1 mole of HCl is 57.62 kJ/mol

The heat liberated by 0.02625 moles of HCl is 0.02625mol x (57.62kJ/mol)

= 1.5125 kJ x1000J/kJ= 1512.5 J

Let t be the final temperature of the solution

Then the amount of heat absorbed by the system , Q = 1512.2 J = mcdt

Where m = mass of solution = 90.0 g

c = specific heat capacity of solution = 4.184 J/goC

dt = change in temperature = final - initial = t - 25

Plug the values we get dt= t-25 = Q/(mc) = 4.02

t = 25+4.02 = 29.02 oC