When 10.0 mL of a 2.0 M acetic acid solution at 20.6 degree C is reacted with 10

Question

When 10.0 mL of a 2.0 M acetic acid solution at 20.6 degree C is reacted with 10.0 mL of a 2.1 M NaOH solution at 21.0 degree C in a coffee cup calorimeter, the resulting temperature of the solution was determined to be 28.6 degree C. Calculate the heat or reaction for this acid-base neutralization. Assume the density and heat capacity of the solution to be the same as water. (d = 1.00 g/ml and C_p = 4.184 J/g degree C) The calorimeter constant was 2.7 J/degree C Select one: a. - 670 J b. -670 J c. - 2500 J d. - 630 J e. - 21 J

Explanation / Answer

moles of acetic acind in 10ml of 2M= Molarity* volume(L)= 2*10/1000 = 0.02

mass of acetic acid =moles* molar mass = 0.02* 60 =1.2

mole of sodium hydroxide in 10ml of 2.1M= 2.1*10/1000=0.021, mass of sodium hydroxide= 0.021*40= 0.84

heat liberates is taken by calorimeter and rise in solution temperature

volume of solution = 10+10= 20ml, density= 1 gm/cc, mass =20*1= 20 gm

heat liberted= heat given to acetic acid+ heat given to sodium hydroxide+ heat taken by calorimeter

= mass of acetic acid* specifc heat* temperature difference + mass of sodium hydroxide* specific heat* temperature difference + calorimeter constant* temperature difference=

10*4.184*(28.6-20.6)+10*4.184*(28.6-21)+2.7*(28.6-21)=673 Joules. close answe is 670 Joules

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