Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 5

Question

Consider two solutions, the first being 50.0 mL of 1.00 M CuSO4 and the second 50.0 mL of 2.00 M KOH . When the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 C to 27.7 C .

From the calorimetric data, calculate H for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL , and that the specific heat and density of the solution after mixing are the same as that of pure water.

Explanation / Answer

Solution:

Calculation of moles of KOH and CuSO4

n KOH = molarity x volume in L

= 2.00 M x 0.050 L

= 0.1 mol

n CuSO4 = 1.00 M x 0.050 L

= 0.050 mol

Neutralization reaction

2 KOH (aq)+ CuSO4 (aq)-- > Cu(OH)2 + K2SO4 (aq)

Calculation heat absorbed by water

q = m C delta T

m is mass of water in g , C is specific heat, Delta T = change in T

q = (100.0 mL x 1 g / mL ) x 4.184 J/g deg C x (27.7-21.5 ) deg C

q = 2594.08 J

Delta H =- q

= -2594.08 J

Delta H = - 2594.08 J

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

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