Determine the enthalpy of neutralization in Joules/mmol for a solution resulting from 19 mL of 0.9 M NaOH solution and 19 mL of a HCl with the same molarity. If separately, each had a temperature of 28.8 degrees Celsius, and upon addition, the highest temperature reached by the solution was graphically determined to be 37.8 degrees Celsius. Round to the nearest whole number.

3.00 J/mmol

Explanation:

The temperature of the solution is increasing, so it's an endothermic reaction. The heat that is being absorbed can be calculated by:

Q = m*c*ΔT

Where m is the total mass (HCl + NaOH), c is the specific heat, and ΔT is the variation of the temperature (final - initial). Because the molarity is small, the solutions are basically water, so c = 4.184 J/g°C.

The mass is the molar mass multiplied by the number of moles (n), which is the volume multiplied by the molarity:

nHCl = 0.9 mol/L * 0.019 = 0.0171 mol

nNaOH = 0.9 mol/L * 0.019 = 0.0171 mol

Molar masses: HCl = 36.5 g/mol; NaOH = 40 g/mol

mHCl = 36.5*0.0171 = 0.62415 g

mNaOH = 40 * 0.0171 = 0.6840 g

m = 1.30815 g

Q = 1.30815*4.184 * (37.8 - 28.8)

Q = 49.2597 J

Because the number of moles of NaOH is equal to the number of moles of HCl, and the stoichiometry of the neutralization is 1:1, they both react completely, so the enthalpy can be calculated based in any of them. The enthalpy is the heat divided by the number of moles. In mmol, n is 17.1 mmol.

ΔH = 49.2597/17.1

ΔH = 2.88 J/mmol

ΔH = 3.00 J/mmol