A chemist uses a coffee-cup calorimeter to completely neutralize 75 ml of 6.67 M HCl with 75 of 6.67 M NaOH. The temperature change of the resulting solution is 39.6 C. Calculate the heat of neutralization of HCl.
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The chemist uses a coffee-cup calorimeter to completely neutralize 75 ml of a solution of HCl with molarity 6.67 mole/liter with 75 ml of a sodium hydroxide solution with a molarity of 6.67 mole/liter.
During the neutralization there is a change in temperature of 39.6 degree Celsius in the temperature of the resulting solution. The heat capacity of water is 4.19 J/g*K. The heat capacity of the resulting solution is also assumed to be the same. As 150 ml of the solution is rising in temperature, the heat being released is 4.19*150 = 628.5 J
The number of moles of HCl in 75 ml of the solution is equal to 6.67*75/1000 = 0.5 mole
The neutralization of 0.5 moles of HCl releases 628.5 J, the heat released when one mole of of HCl is neutralized is 1257 J/mole or 1.257 kJ/mole.
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