Question

If 200 mL of aqueous 0.862 M HCl is mixed with 20. mL of aqueous 0.431 M Ba(OH)2 in a constant-pressure calorimeter of negligible heat capacity, the initial temperature of both solutions is the same at 20.48ºC. The neutralization reaction H+(aq) + OH–(aq) → H2O(l) has a ΔHº = –56.2 kJ/mol.

What is the final temperature of the mixed solution? (Assume all solutions have a density of 1.00 g/cm3 and a specific heat capacity of 4.184 J/g•ºC)

Answer 1

To find the final temperature of the mixed solution, we can use the equation q = mcΔT for both the HCl and Ba(OH)2 solutions. By setting the heat absorbed by the HCl equal to the heat released by the Ba(OH)2 and solving the equation, we can find the final temperature.

Step-by-step explanation:

To solve this problem, we can use the equation q = mcΔT, where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. First, we need to calculate the heat absorbed by the HCl solution and the heat released by the Ba(OH)2 solution. We can use the equation q = nΔH, where n is the number of moles and ΔH is the enthalpy change. The heat absorbed by HCl can be calculated by multiplying the number of moles of HCl by its enthalpy change (-56.2 kJ/mol). The heat released by Ba(OH)2 can be calculated by multiplying the number of moles of Ba(OH)2 by its enthalpy change (0 kJ/mol). Since the reaction is exothermic and heat is released, the heat absorbed by HCl is equal to the heat released by Ba(OH)2. We can set up an equation and use algebra to solve for the final temperature. The equation is: q(HCl) = q(Ba(OH)2) M(HCl) * c * ΔT(HCl) = M(Ba(OH)2) * c * ΔT(Ba(OH)2) 200 mL * 0.862 M * 4.184 J/g•ºC * (T - 20.48ºC) = 20 mL * 0.431 M * 4.184 J/g•ºC * (T - 20.48ºC) Solving this equation will give us the final temperature of the mixed solution.

Answer 2

The final temperature of the mixed solution is 26.26 °C.

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