If 50.00mL of 1.10M NaOH is added to 25.00mL of 1.86M HCl, with both solutions originally at 24.70 degrees Celsius, what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02g/mL and a specific heat of 3.98J/g*degrees Celcius.)
Express answer in four significant figures.
3 answers
You have a great deal of information but you're still missing the heat of reaction (heat of neutralization).
oops, here it is:
-55.84 kJ/mol H20 produced
-55.84 kJ/mol H20 produced
50.00 x 1.10 = 55.00 millimols NaOH
25.00 x 1.86 = 46.5 mmols HCl
.......NaOH + HCl ==> NaCl + H2O
I......55.00..46.50.....0......0
C.....-46.5..-46.5.....46.5...46.5
E......8.5.....0.......46.5...46.5
That 46.5 millimols = 0.0465 mols H2O.
55.84 kJ/mol x 0.0465 mols = 2.597 kJ heat = q.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
I would change q to J from kJ, use density to find mass of the 75.00 mL H2O, substitute Ti and sp.h. and solve for Tf. (I am ignoring the specific heat of the unreacted NaOH(solid) as well as that of the NaCl(solid) formed.)
25.00 x 1.86 = 46.5 mmols HCl
.......NaOH + HCl ==> NaCl + H2O
I......55.00..46.50.....0......0
C.....-46.5..-46.5.....46.5...46.5
E......8.5.....0.......46.5...46.5
That 46.5 millimols = 0.0465 mols H2O.
55.84 kJ/mol x 0.0465 mols = 2.597 kJ heat = q.
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
I would change q to J from kJ, use density to find mass of the 75.00 mL H2O, substitute Ti and sp.h. and solve for Tf. (I am ignoring the specific heat of the unreacted NaOH(solid) as well as that of the NaCl(solid) formed.)
2 answers