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Question

Calculate the enthalpy change when 50 mL of 0.01 M Ca(OH)2 reacts with 25 mL of 0.01 M HCl. Given that ΔHo of neutralization of a strong acid and a strong base is 14 kcal mol1

A
6.4 cal
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B
3.5 cal
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C
1.4 cal
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D
9.4 cal
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Solution

The correct option is B 3.5 cal
First we will calculate the no. of moles:
Number of moles of HCl=Molarity×Volume1000=0.01×251000=25×105

Now HCl dissociates as
HClH++Cl

So, nH+=nHCl=25×105

Number of moles of Ca(OH)2=Molarity×Volume1000=0.01×501000=50×105
nOH=2×nCa(OH)2=2×50×105=100×105

In the process of neutralisation 25×105 mole H+ will be completely neutralised since it is the limiting reagent
ΔH= Molar enthalpy of neutralisation × Number of moles
=14×25×105kcal=0.0035 kcal=3.5 cal

Theory:

Standard Enthalpy of Neutralization ΔneutHo

Heat released when one mole of H+ in dilute solution combines with one mole of OH to give rise to undissociated water at 1 bar, 298 K.

The amount of heat released in formation of one mole of water when an acid is neutralised by a base.





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