A 21.3-mL sample of 0.977 M NaOH is mixed with 29.5 mL of 0.918 M HC1 in a coffe

Question

A 21.3-mL sample of 0.977 M NaOH is mixed with 29.5 mL of 0.918 M HC1 in a coffee-cup calorimeter (see Section 6.6 of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest whole-number coefficients, is -55.8 kJ. Both solutions are at 19.6 degree C prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is 1.00 g/mL, the specific heat of all solutions is the same as that of water, and volumes are additive.

Explanation / Answer

mmol of NaOH = M*V = 0.977 M * 21.3 mL = 20.81 mmol

mmol of HCl = M*V = 0.918 M * 29.5 mL = 27.08 mmol

20.81 mmol of both will react that is 0.02081 mol of each reacted

reaction is:

HCl + NaOH —> NaCl + H2O H = -55.8 KJ

So,

heat released = number of mol * H

= 0.02081 * 55.8

= 1.161 KJ

= 1161 J

Total volume of solution, V = 21.3 mL + 29.5 mL = 50.8 mL

Since density is 1 g/mL, mass of solution = 50.8 g

specific heat capacity, C = 4.184 J/g.oC

now use:

Q = m*C*(Tf - Ti)

1161 J = 50.8 g * 4.184 J/g.oC*(Tf - 19.6) oC

Tf - 19.6 = 5.46

Tf = 25.1 oC

Answer: 25.1 oC

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