A 71.0 mL sample of 1.0 M NaOH is mixed with 49.0 mL of 1.0 M H_2SO_4 in a large

Question

A 71.0 mL sample of 1.0 M NaOH is mixed with 49.0 mL of 1.0 M H_2SO_4 in a large Styrofoam coffee cup; the cud is fitted with a lid through which passes a calibrated thermometer. The temperature each solution before mixing is 23.9 degree C. After adding the NaOH solution to the coffee cup, the mixed solutions are stirred until reaction is complete. Assume that the density of the mixed solutions is 1.00 g/mL, that the specific heat of the mixed solutions is 4.18 J/(g middot degree C), and that no heat is lost to the surroundings. The Delta H_rxn for the neutralization of NaOH with H_2SO_4 is - 114 kJ/mol H_2SO_4. What is the maximum measured temperature in the Styrofoam cup?

Explanation / Answer

H2SO4 + 2 NaOH Na2SO4 + 2 H2O
Because both solutions have the same concentrations:
71.0 ml of NaOH would react completely with 71.0 x (1/2) = 35.5 ml of H2SO4, but there is more H2SO4 present than that, so NaOH is the limiting reactant and H2SO4 is in excess.

(0.0710 L) x (1.0 mol/L NaOH) x (1 mol H2SO4 / 2 mol NaOH) x (114 kJ/mol H2SO4) = 4.047 kJ

(4047 J) / (4.18 J/(g °C)) / ((71.0 ml + 49.0 ml) x (1.00 g/ml)) = 8.06 °C change

(23.9 °C + 8.06 °C) = 31.96 °C

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