Answer:
22.48°C is the final temperature of the solution.
Explanation:
Heat of neutralization of reaction , when 1 mol of nitric acid reacts= ΔH= -56.2 kJ/mol= -56200 J/mol
![Moles (n)=Molarity(M)\times Volume (L)](https://tex.z-dn.net/?f=Moles%20%28n%29%3DMolarity%28M%29%5Ctimes%20Volume%20%28L%29)
Moles of nitric acid = n
Volume of nitric acid solution = ![8.10\times 10^2 mL= 0.81L](https://tex.z-dn.net/?f=8.10%5Ctimes%2010%5E2%20mL%3D%200.81L)
Molarity of the nitric acid = 0.600 M
![n=0.600 M\times 0.81 L=0.486 mol](https://tex.z-dn.net/?f=n%3D0.600%20M%5Ctimes%200.81%20L%3D0.486%20mol)
Moles of barium hydroxide = n'
Volume of barium hydroxide solution = ![8.10\times 10^2 mL= 0.81L](https://tex.z-dn.net/?f=8.10%5Ctimes%2010%5E2%20mL%3D%200.81L)
Molarity of the barium hydroxide= 0.300 M
![n'=0.300 M\times 0.81 L=0.243 mol](https://tex.z-dn.net/?f=n%27%3D0.300%20M%5Ctimes%200.81%20L%3D0.243%20mol)
![2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_3+2H_2O](https://tex.z-dn.net/?f=2HNO_3%2BBa%28OH%29_2%5Crightarrow%20Ba%28NO_3%29_3%2B2H_2O)
According to reaction, 2 mol of nitric acid reacts with 1 mol of barium hydroxide .Then 0.486 mol of nitric acid will react with :
barium hydroxide.
Heat release when 0.486 mol of nitric acid reacted = Q
= ΔH × 0.486 = -56200 J/mol × 0.486 mol=-27,313.2 J
Heat absorbed by the mixture after reaction = Q' = -Q = 27,313.2 J
Volume of the nitric solution = 0.81 L = 810 mL
Volume of the ferric nitrate solution = 0.81 L = 810 mL
Total volume of the solution = 810 mL + 810 mL = 1620 mL
Mass of the final solution = m
Density of water = density of the final solution = d = 1 g/mL
![Mass=density\times Volume](https://tex.z-dn.net/?f=Mass%3Ddensity%5Ctimes%20Volume)
![m=1 g/ml\times 1620 ml=1620 g](https://tex.z-dn.net/?f=m%3D1%20g%2Fml%5Ctimes%201620%20ml%3D1620%20g)
Initial temperature of the both solution were same = ![T_1=18.46^oC](https://tex.z-dn.net/?f=T_1%3D18.46%5EoC)
Final temperature of the both solution will also be same after mixing= ![T_2](https://tex.z-dn.net/?f=T_2)
Heat capacity of the mixture = c = 4.184 J/g°C
Change in temperature of the mixture = ΔT =![(T_2-T_1)](https://tex.z-dn.net/?f=%20%28T_2-T_1%29)
![Q=mc\Delta T=mc(T_2-T_1)](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T%3Dmc%28T_2-T_1%29)
![27,313.2 J= 1620 g\times 4.184 J/g^oC\times (T_2-18.46^oC)](https://tex.z-dn.net/?f=%2027%2C313.2%20J%3D%201620%20g%5Ctimes%204.184%20J%2Fg%5EoC%5Ctimes%20%28T_2-18.46%5EoC%29)
![T_2=22.49^oC](https://tex.z-dn.net/?f=T_2%3D22.49%5EoC)
22.48°C is the final temperature of the solution.