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If 50.00 mL of 1.05 M sodium hydroxide is added to 25.00 mL of 1.88 M hydrochloric acid, with both solutions originally at 24.66°C, 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 1.02 g/mL and a specific heat of 3.98 Jg⁻¹°C⁻¹.)

2 Answers

3 votes

Final answer:

To find the final solution temperature, we need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other. By plugging in the given values and solving the equation, we find that the final solution temperature will be 24.77 °C.

Step-by-step explanation:

To find the final solution temperature, we can use the principle that the heat given off by the reaction is equal to that taken in by the solution. We need to calculate the heat exchanged by the reaction and the heat exchanged by the solution and set them equal to each other.

First, we calculate the heat exchanged by the reaction using the equation:

q_reaction = C_reaction * ΔT_reaction

where C_reaction is the heat capacity of the reaction solution and ΔT_reaction is the change in temperature of the reaction.

Next, we calculate the heat exchanged by the solution using the equation:

q_solution = m_solution * C_solution * ΔT_solution

where m_solution is the mass of the solution, C_solution is the specific heat of the solution, and ΔT_solution is the change in temperature of the solution.

Now we can set the two heat exchanges equal to each other and solve for the final solution temperature:

q_reaction = q_solution

C_reaction * ΔT_reaction = m_solution * C_solution * ΔT_solution

Plugging in the given values:

C_reaction = C_solution = 3.98 Jg⁻¹°C⁻¹

m_solution = (50.00 mL of NaOH * 1.02 g/mL) + (25.00 mL of HCl * 1.02 g/mL) = 76.50 g

ΔT_reaction = (28.9 °C - 24.66 °C) = 4.24 °C

ΔT_solution = ?

Now we can solve for ΔT_solution:

3.98 Jg⁻¹°C⁻¹ * 4.24 °C = 76.50 g * 3.98 Jg⁻¹°C⁻¹ * ΔT_solution

ΔT_solution = (3.98 Jg⁻¹°C⁻¹ * 4.24 °C) / (76.50 g * 3.98 Jg⁻¹°C⁻¹) = 0.1107 °C

Finally, we calculate the final solution temperature:

Final Temperature = 24.66 °C + 0.1107 °C = 24.77 °C

User Petelids
by
6.4k points
1 vote

Answer:

The final temperature of solution is 24.668883°C

Step-by-step explanation:

The enthalphy change for the reaction of 1 mol of strong acid and strong base is -57.62 J/mol.

moles of NaOH = 0.0525 mol

moles of HCl = 0.047 mol

Hence ,

Heat released in this reaction is .047 × 57.62 joules

= 2.70814 joules

Final mass of solution is density × final volume.

= 1.02 × 75

=76.5 grams

And ,

MS = 76.6 × 3.98

= 304.868

Since,

Q = M × s × ΔT

Therefore,

ΔT =
(Q)/(MS)

=
(2.70814)/(304.868)

= 0.00888299198 °C

Hence ,

Final temperatre = 24.66 + .00888299198

= 24.668883 °C

User TamerB
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7.0k points