A sample of 25.00 mL of HCl(aq) was mixed with 25.00 mL of KOH(aq) of the same c

Question

A sample of 25.00 mL of HCl(aq) was mixed with 25.00 mL of KOH(aq) of the same concentration in a calorimeter. At a result, the temperature rose from 25.00 to 26.60 degree C. Given that H degree = -56 02 kJ/mol for the reaction and that the specific heat of dilute aqueous solutions is approximately equal to that of water (4.184 J/g degree C), determine the molarity of the original HC1 solution. (Assume the heat capacity of the calorimeter to be negligible.). H_3O^+(aq) + OH (aq) rightarrow H_2O(l) 0.24 M 0.48 M 0.024 M 0.0060 M 0.012 M

Explanation / Answer

Let the number of moles of HCl solution added = m

Since both are strong acid and base, moles of HCl = moles of NaOH

amount of heat released = 56.02 *10^3 *m J

Heat absorbed by the calorimeter is negligible as the heat capacity of calorimeter is assumed to be negligible. So, the heat released due to the neutralization reaction will be used to increase the temperature of the water.

Total volume of the water (solution) = 25 + 25 = 50 mL

Since density of water 1gm/ mL, So, 50mL water = 50 gm water.

heat absorbed by the water = 50 gm * 4.18 J/gm/oC *(26.60-25)

50 gm * 4.18 J/gm/oC *(26.60-25) = 56.02 *10^3 *m J

or, m =5.969*10^-3 moles

25 mL solution contain 5.969*10^-3 moles of acid. Molarity of acid = 5.969*10^-3 moles*1000/25 mL

= 0.238 moles/L = 0.24 M

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