A student carefully measures out 200.0 mL of an aqueous solution of 1.0 M HCl in

Question

A student carefully measures out 200.0 mL of an aqueous solution of 1.0 M HCl in a coffee cup calorimeter.  She then carefully adds 200.0 mL of 1.0 M NaOH to the calorimeter.  The initial temperature for both solutions is 20.0 A student carefully measures out 200.0 mL of an aqueous solution of 1.0 M HCl in a coffee cup calorimeter.  She then carefully adds 200.0 mL of 1.0 M NaOH to the calorimeter.  The initial temperature for both solutions is 20.0 A student carefully measures out 200.0 mL of an aqueous solution of 1.0 M HCl in a coffee cup calorimeter.  She then carefully adds 200.0 mL of 1.0 M NaOH to the calorimeter.  The initial temperature for both solutions is 20.0

Explanation / Answer

You need to calculate how many moles of HCl is neutralizing

moles of HCl = 1 * 200/1000 = 0.2 mol

As

HCl(aq)+ NaOH(aq) ?NaCl(aq)+H2O(l) ?H = -55.0 kJ/mol

Therefore heat relased for 0.2 mole HCl neutralization = 0.2 mol * 55 kJ/mol

11 kJ = 11 * 1000 J = 11000 J (Note heat is absorbed by the water so I have kept it positive, with respect to water)

volume of solution = (200 + 200) mL = 400 mL

as density of solution = density of water = 1 g/mL

therefore mass of solution = 400 ml * 1 g/mL = 400 g

heat capacity of solution = heat capacity of water = 4.184 J / g C

T-final -T-initial = heat absorbed / (mass of solution * heat capacity of solution)

T-final - 20 C = 11000 / (400 * 4.184)

T- final - 20 C = 6.6 C

T-final = 20 C + 6.6 C = 26.6 C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.