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Q.

The enthalpy changes when 50ml 0.01M CaOH2 reacts with 25ml  0.01M HCl  is  5x+2cal. Given that ΔH0 neutralization of strong acid and strong base is 140Kcal/mole. x  is____________

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answer is 5.

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Detailed Solution

Equation H+=50×0.021000=10−3eqH+

Eq.OH−=25×0.011000=25×10−5OH−

∴25×10−5eq.H2O  forms

⇒25×10−5×140=0.035Kcal

35cal

∴55+2Cal

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