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The enthalpy changes when 50ml 0.01M $C a {(O H)}_{2}$ reacts with 25ml 0.01M $H C l$ is $5 (x + 2) c a l$ . Given that $Δ H^{0}$ 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 $H^{+} = \frac{50 \times 0.02}{1000} = 10^{- 3} e q H^{+}$

$E q . O H^{-} = \frac{25 \times 0.01}{1000} = 25 \times 10^{- 5} O H^{-}$

$∴ 25 \times 10^{- 5} e q . H_{2} O f o r m s$

$\Rightarrow 25 \times 10^{- 5} \times 140 = 0.035 K c a l$

$35 c a l$

$∴ 5 (5 + 2) C a l$

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