[billyboy]

Nitric acid reacts with potassium hydroxide as follows:

HNO3 (aq) + KOH(aq) --> KNO3 (aq) + H2O(l)

In an experiment a student placed 55.0 mL of 1.3 mol/L HNO3 and 55.0 mL of 1.3 mol/L KOH into a
calorimeter at 23.5 ̊C. The temperature rose to 31.8 ̊C after the reaction was complete. Calculate the heat
of reaction (∆HR) in kJ/mol of acid. Assume that the specific heat capacity and density of both solutions
are 4.2 J/g ̊C and 1 g/mL respectively.

I've been stuck on this question for a while, I just keep getting the wrong answer.

[Borek]

Show how you got the wrong answer and we will start from there.

[billyboy]

c= 4.2 j/g*C
deltaT = 31.8*C - 23.5*C
          = 8.3*C

mass of HNO3 = 55.0mL/1 x 1g/1mL
                      = 55.0g

Q= mcDeltaT
  = (55.0g)(4.2j/g*C)(8.3*C)
  = 1917 x 1kj/1000j
  = 1.917kj

n= m/M
  = 55.0g/63.02g/mol
  = 0.873mol

DeltaH= -Q/n
          = -(1.917kj)/0.873mol
          = -2.20kj/mol

But the answer should be -54kj/mol

[Orcio_87]

@billyboy

Q= mcDeltaT
  = (55.0g)(4.2j/g*C)(8.3*C)
  = 1917 x 1kj/1000j
  = 1.917kj

Volume (mass) of solution is greater than that!

n= m/M
  = 55.0g/63.02g/mol
  = 0.873mol]

Concentration (and quantity) of HNO₃ is given text!

[billyboy]

I'm still confused because the mass of HNO3 in my mind can't be higher than 55.0g because it's a 1:1 ratio between mL and grams in this question.

Also with the concentration, meaning 1.3mol/L, I still get the wrong answer:

55.0mL x 1L/1000mL
= 0.055L

1.3mol x 0.055L
= 0.0715mol

DeltaH= -Q/n
          = - (1.917kj) / 0.0715mol
          = 26.81kj/mol

[Orcio_87]

Also with the concentration, meaning 1.3mol/L, I still get the wrong answer.. 26,81 kJ/mol

Right, but only because of the volume (and then - heat) is greater than yours calculation.

(there are two miscible solutions, not only one)

[billyboy]

Do you mean that the mass should be 110g instead of 55.0g? Even so, I still got the wrong answer again with my final answer being -26.82kj/mol.

[Borek]

mass of HNO3 = 55.0mL/1 x 1g/1mL
                      = 55.0g

This is mass of the acid solution, not of the acid itself.

But you got the correct number of moles later. Note that the question is made easy by giving identical volumes of concentrations of the acid and the base. It is not unusual to have to start with the limiting reagent calculations.

Q= mcDeltaT
  = (55.0g)(4.2j/g*C)(8.3*C)
  = 1917 x 1kj/1000j
  = 1.917kj

After mixing solutions you end end with 55mL+55mL=110mL and it is this final volume that heats up.

Do you mean that the mass should be 110g instead of 55.0g?

Yes, see above.

Even so, I still got the wrong answer again with my final answer being -26.82kj/mol.

That is exactly half of the correct answer, so you are still doing something wrong, check your math, as you have all correct data now.