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)`

Options

A. 14 kcal
B. 35 cal
C. 10 cal
D. 7.5 cal

Answer

B

Explanation

Number of moles of `HCl=(MV)/(1000)=(0.01xx25)/(1000)`
`=25xx10^(-5)`
`HCl to H^(+)+Cl^(-)`
`n_(H^(+))=25xx10^(-5)`
Number of moles of `Ca(OH)_(2)=(MV)/(1000)=(0.01xx50)/(1000)=50xx10^(-5)`
`n_(OH^(-))=2xx50xx10^(-5)=10^(-3)`
In the process of neutralisation `25xx10^(-5)` mole `H^(+)` will be completely neutralised.
`thereforeDeltaH=140xx25xx10^(-5)kcal` `=0.035kcal=35cal`