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InterviewSolution
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Calculate the enthalpy change when `50 mL` of `0.01 M Ca(OH)_(2)` reacts with `25mL` of `0.01 M HCI`. Given that `DeltaH^(Theta)` neutralisaiton of strong acid and string base is `140 kcal mol^(-1)`A. `14 kcal`B. `35 cal`C. `10 cal`D. `7.5 cal` |
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Answer» Number of moles of `HCI = (MV)/(1000) = (0.01xx25)/(1000)` `= 25 xx 10^(-5)` `HCI rarr H^(o+) + CI^(Theta)` `n_(H)^(o+) = 25 xx 10^(-5)` Number of moles of `Ca(OH)_(2) = (MV)/(1000) = (0.01xx50)/(1000) = 50 xx 10^(-5)` `Ca(OH)_(2) hArr Ca^(2+) +2 overset(Theta)OH` `n_(OH)^(Theta) = 2 xc 50 xx 10^(-5) = 10^(-3)` In the process of neutralisation, `25 xx 10^(-5) "mole" H^(o+)` will be completely neutralised. `:. DeltaH^(Theta) = 140 xx 25 xx 10^(-5) kcal = 0.035 kcal = 35 cal` |
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