A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HCl

Question

A 100.0 mL sample of 0.300 M NaOH is mixed with a 100.0 mL sample of 0.300 M HCl in a coffee cup calorimeter. If both solutions were initially at 35.00°C and the temperature of the resulting solution was recorded as 37.00°C, determine the H°rxn (in units of kJ/mol NaOH) for the neutralization reaction between aqueous NaOH and HCl. Assume (1) that no heat is lost to the calorimeter or the surroundings, and (2) that the density and the heat capacity (4.18 J/g oC)of the resulting solution are the same as water.

Explanation / Answer

Answer – We are given volume of HCl = 100 mL , [HCl] = 0.300 M

Volume of NaOH = 100 mL , [NaOH] = 0.300 M , ti = 35.00 oC , tf = 37.00 oC.

Total volume = 100 mL+ 100 mL = 200 mL, density of water = 1.0 g/mL

So, mass of water = 200 g

Reaction - HCl + NaOH -----> NaCl + H2O

We know the heat formula

q = m*C* t

    = 200 g * 4.184 J/goC*(37.00 – 35.00)oC

    = 1673.6 J

   = 1.674 kJ

We know the in the neutralization reaction heat gets evolved, so

Horxn = -q

          = -1.674 kJ

Now we need to calculate the Horxn per moles of NaOH

Moles of NaOH = 0.300 M * 0.100 L

                       = 0.0300 moles

Horxn per moles of NaOH = -1.674 kJ / 0.0300 moles

                                      = - 55.8 kJ/mol

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