In a titration process, a 10 mol% aqueous sulfuric acid solution (specific gravi

Question

In a titration process, a 10 mol% aqueous sulfuric acid solution (specific gravity of 1.27) is neutralized with a 3-molar sodium hydroxide solution (specific gravity of 1.13) according to the following reaction:

H2SO4 (aq) + 2NaOH (aq) Na2SO4 (aq) + 2H2O (l)

a) Calculate the volume ratio between the sodium hydroxide solution to the sulfuric acid solution.

b) Calculate the amount of heat released during the titration process if both the reactants and products are at 25°C. Note: the heat of solution of Na2SO4 is -1.17 kJ/mol.

Explanation / Answer

H2SO4 (aq) + 2NaOH (aq) ----> Na2SO4 (aq) + 2H2O (l)
M1V1/n1 = M2V2/n2
10*V1/1 = 3*V2/2
V1/V2 = 3/20
V1 : V2 = 3 : 20 ratio
H2SO4 (aq) + 2NaOH (aq) -----> Na2SO4 (aq) + 2H2O (l)
It is a neutralisation reaction between strong acid and strong base then dH change is always -13.5 k Cal or -58.5 kJ for 1 mole formation of water.
here 2 moles of H2O is formed then dH of neutralisation = 2*(-58.5) = -117 kJ
Heat of solution Na2SO4 = -1.17 kJ per mol
from the reaction dH for Product = -117 + (-1.17) = -118.17 kJ

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