About 18.5 mL of 0.10 M HCl was reacted with 13.0 mL of 0.14 M NaOH. Each solution was originally at 25.0 °C, and then is thoroughly mixed. The final temperature of the neutralized mixture is 29.7 °C. Find the change in enthalpy (kJ/mol HCl) for the reaction.
Neutralization reaction is
NaOH + HCl ..................> NaCl + H2O
18.5 mL of 0.10 M HCl = 0.0185 L * 0.10 mole / L = 0.00185 mole.
13.0 mL of 0.14 M NaOH. = 0.013 L * 0.14 mole / L = 0.00182 mole.
total volume of solution = (18.5 + 13.0) = 31.5 ml
assuming density of solution = 1.0 g / ml
thus
mass of solution = 31.5 g
heat = m * s* dT = 31.5 * 4.184 * (29.7 - 25.0) = 619.4 J = 0.6194 KJ
change in enthalpy (kJ/mol HCl) for the reaction = - 0.6194 KJ / 0.00185 mol = - 334.8 KJ / mole. (answer)
About 18.5 mL of 0.10 M HCl was reacted with 13.0 mL of 0.14 M NaOH....