Heat of neutralisation of HF (a weak acid) with strong base is - text 16.4 text kcal. Calculate heat of ionisation of HF in water.

Question

Heat of neutralisation of HF (a weak acid) with strong base is - text 16.4 text kcal. Calculate heat of ionisation of HF in water. (A)  -13.7 text kcal  (B)  -2.7 text kcal  (C)  +30.1 text kcal  (D)  + text 3.01 text kcal

Tardigrade Admin · Accepted Answer

(i) HF+OH- xrightarrowH2O+F-;Δ H =-16.4 kcal (ii) H++OH- xrightarrowH2O;Δ H=-13.7 kcal From Eqs. (i)-(ii) gives the required equation (ionisation of HF in water) HF xrightarrowH++F-;Δ H =-2.7kcal.

Q. Heat of neutralisation of HF (a weak acid) with strong base is $- 16.4 k c a l .$ Calculate heat of ionisation of HF in water.

$- 13.7 k c a l$

$- 2.7 k c a l$

$+ 30.1 k c a l$

$+ 3.01 k c a l$

(i) $H F + O H^{-} H_{2} O + F^{-}; Δ H = - 16.4 k c a l$
(ii) $H^{+} + O H^{-} H_{2} O; Δ H = - 13.7 k c a l$
From Eqs. (i)-(ii) gives the required equation (ionisation of HF in water)
$H F H^{+} + F^{-}; Δ H = - 2.7 k c a l .$

Q. Heat of neutralisation of HF (a weak acid) with strong base is −16.4kcal. Calculate heat of ionisation of HF in water.

−13.7kcal

−2.7kcal

+30.1kcal

+3.01kcal