What is the final temperature when 50.0 mL of 0.100 M hydrochloric acid solution at 20.0 oC is neutralized by 50.0 mL of 0.100 M sodium hydroxide solution at 20.0 oC ?

Answer 1

For this problem we need the ∆H for the reaction, which isn't provided, but we will use the commonly accepted value of -55.90 kJ/mol for the neutralization of a strong acid and a strong base.

HCl + NaOH ==> NaCl + H2O ... balanced equation

mols HCl = 50.0 ml x 1 L/1000 ml x 0.100 mol/L = 0.005 mols HCl

mols NaOH = 50.0 ml x 1L/1000 ml x 0.100 mol/L = 0.005 mols NaOH

∆Hrxn = 0.005 mols x -55.90 kJ/mol = -0.2795 kJ of heat released from the system to the surroundings.

Since this is an exothermic reaction (heat given off), the temperature of the solution will INCREASE.

We now use q = mC∆T

q = heat = 0.2795 kJ = 279.5 J

m = mass = 50 ml + 50 ml = 100 mls x 1g/ml = 100 g (we assume a density of 1 g/ml)

C = specific heat = 4.814 J/gº

∆T = change in temperature = ?

∆T = q / mC = 279.5 J / (100 g)(4.184 J/gº)

∆T = 0.668º

Final temperature = 20º + 0.668º = 20.7ºC