The enthalpy of formation of `H_(2)O(l)` is -280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong base is -56.70 kJ/mol. What is the enthalpy of formation of `OH^(-)` ions?

Options

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

Explanation

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`