Question

3) You would like to measure the heat of neutralization of an acid with a base. You mix 225 mL each of 0.65 M HNO, and 0.65 M LiOH, both at a temperature of 18.1 °C in a calorimeter cup equilibrated to that same temperature. After following the temperature change for 10 min and extrapolating it back to the time of addition, you find that the "final temperature" after mixing was 22.3 °C. Previously, you measured the heat capacity of the calorimeter cup (Ceup) to be 47 JPC Assuming that the density of the two solutions is 1.00 g/ml and that their specific heats are the same as water, 4.184 J/g C, What is the heat absorbed by the two solutions, sola? What is the heat absorbed by the calorimeter cup, gcup? What is the heat change for the neutralization reaction, gra? What is the AH for the neutralization? Write the net ionic equation for this neutralization reaction, including the states of matter (i.e., H'aą) for the reactants and products.

Answer 1

the balanced reaction

HNO3(aq) + LiOH(aq) = H2O(l) + LiNO3(aq)

Part a

Mass of solution = volume x density

= (225 mL x 2) x 1 g/mL = 450 g

Heat absorbed by the solution

qsoln = mass x Cp x (T2-T1)

= 450 g x 4.184 J/gC x (22.3 - 18.1)C

= 7907.76 J

Part b

Heat absorbed by calorimeter

q cup = C cup x (T2-T1)

= 47 J/C x (22.3 - 18.1)C

= 197.4 J

Part c

heat released by reaction

qrxn = qsoln + q cup

= 7907.76 + 197.4

= 8105.16 J

Part d

Moles of HNO3 = molarity x volume

= 0.65 mol x 225 mL x 1L/1000 mL

= 0.14625 mol

Moles of LiOH = = 0.65 mol x 225 mL x 1L/1000 mL

= 0.14625 mol

ΔH = (qrxn / mol)

= 8105.16 J / 0.14625 mol

= 55419.90 J/mol x 1kJ/1000 J

= 55.42 kJ/mol

Part e
The balanced reaction

HNO3(aq) + LiOH(aq) = H2O(l) + LiNO3(aq)

Total ionic equation

H+(aq) + NO3-(aq) + Li+(aq) + OH-(aq) = H2O(l) + Li+(aq) + NO3-(aq)

Net ionic equation

H+(aq) + OH-(aq) = H2O(l)