60.0 mL of 0.350 M H2S04 solution is mixed with 60.0 mL of 0.700 M NaOH solution

Question

60.0 mL of 0.350 M H2S04 solution is mixed with 60.0 mL of 0.700 M NaOH solution in a coffee cup calorimeter of negligible capacity. The initial temperature of the two solutions are both at 23.0 degrees C. What is the final temperature? (What is the temperature change?) assume solutions have same specific heat and density as water. ( s water = 4.184 J/goc; d H20-1.00g/mL) H2SO4 + 2NaOH"> Na2SO4 + 2H2O Hrxn=-114 kJ/mol rxn 18.9 degrees C 27.8 degrees 22.5 degrees 26.3 degrees C 30.1 degrees C 23.3 degrees C 24.9 degrees C 28.6 degrees C

Explanation / Answer

mol of base = MV = 0.7*60 = 42 or 0.042 mol

mol of acid = 0.042/2 = 0.021 mol

Qsoln = m*C*(Tf-Ti)

m = 60 mL + 60 mL = 120 mL = 120 g

Qsoln = (120*4.184)(Tf-23)

Qsoln = -Qrxn

Qrxn = n*HRxn = 0.021*-114 = -2.394 kJ

Qsoln = 2.394 = 2394 J

Qsoln = (120*4.184)(Tf-23)

2394 = (120*4.184)(Tf-23)

Tf = 2394 /(120*4.184) + 23

Tf = 27.768 °C

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