In a coffee-cup calorimeter, 150.0 ml of 0.50 m hcl is added to 50.0 ml of 1.00 m naoh to make 200.0 g solution at an initial temperature of 48.2°c. if the enthalpy of neutralization for the reaction between a strong acid and a strong base is −56 kj/mol, calculate the final temperature of the calorimeter contents. assume the specific heat capacity of the solution is 4.184 j°c⁻¹ g⁻¹ and assume no heat loss to the surroundings.

51.54°C the final temperature of the calorimeter contents.

Explanation:

Molarity of HCl= 0.50 M

Volume of HCl= 150.0 mL = 0.150 L

Moles of HCl= n

Molarity of NaOH= 1.00 M

Volume of NaOH= 50.0 mL = 0.050 L

Moles of NaOH= n'

Since moles of NaOH are less than than moles of HCl. so energy release will be for neutralization of 0.050 moles of naOH by 0.050 moles of HCl.

n = 0.050

(1 kJ= 1000 J)

The energy change released during the reaction = 2800 J

Volume of solution = 150.0 mL + 50.0 mL = 200.0 mL

Density of the solution (water) = 1.00g/mL

Mass of the solution , m= 200 mL × 1.00 g/mL = 200 g

Now , calculate the final temperature by the solution from :

where,

q = heat gained = 2800 J

c = specific heat of solution =

= final temperature =

= initial temperature =

Now put all the given values in the above formula, we get:

51.54°C the final temperature of the calorimeter contents.

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