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50 mL of 1M hydrochloric acid with 50 mL of 1M sodium hydroxide produces a tempe

ID: 856789 • Letter: 5

Question

50 mL of 1M hydrochloric acid with 50 mL of 1M sodium hydroxide produces a temperature increase of 6.7 oC.

1) Use the total solution volume, density of water, measured temperature change, and specific heat of water to calculate the heat absorbed by a solution (qsoln) mixed in the proportions described by the response regulations.

2) Calculate the amount of heat that would be released during an acid-base neutralization reaction (qneut),

3) Calculate the number of moles of H+ and OH- intially present and the number of moles of H2O produced.

50 mL of 1M hydrochloric acid with 50 mL of 1M sodium hydroxide produces a temperature increase of 6.7 degree C. 1) Use the total solution volume, density of water, measured temperature change, and specific heat of water to calculate the heat absorbed by a solution (qsoln) mixed in the proportions described by the response regulations. 2) Calculate the amount of heat that would be released during an acid-base neutralization reaction (qneut), 3) Calculate the number of moles of H+ and OH- intially present and the number of moles of H2O produced. 4) Calculate delta Hneut per mole of H2O produced.

Explanation / Answer

1) Total volume of solution = 50 + 50 = 100mL

Density of water = 1 g/mL

Mass of the solution(m) = density x volume = 1 x 100 = 100 g

specific heat of water(S) = 4.185 J/g.K

qsol = m x S x dT = 100 x 4.185 x 6.7 = 2803.95 J.

2) HCl + NaOH -------------> NaCl + H2O ; dH = -55.8kJ/mol

No.of moles of HCl = conc.x volume = 1 x 50/1000 = 0.05 moles

No.of moles of NaOH = 1 x 50/1000 = 0.05 moles

From the equation, The standard enthalpy change of neutralization for a strong acid and base is -55.8 kJ/mol,

qneut = -55.8 kJ/mol x 0.05 mol = -2.79 kJ

3) No.of moles of H+ = 1 x 50/1000 = 0.05 moles

No.of moles of OH- = 1 x 50/1000 = 0.05 moles

No.of moles of H2O formed = 0.05 moles

4) dH = q/n = 2803.95 / 0.05 = 56079 J/mol = 56.079 kJ/mol

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