When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous solution, 106 kJ of heat is liberated. Predict the enthalpy of ionization of the acid ?

$4.3 k J m o l^{- 1}$

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$- 4.3 k J m o l^{- 1}$

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$- 8.6 k J m o l^{- 1}$

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$8.6 k J m o l^{- 1}$

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Solution

The correct option is D
$8.6 k J m o l^{- 1}$

$H_{2} C_{2} O_{4 (a q)} \to 2 H_{a q}^{+} + C_{2} O_{4 (a q)}^{2 -}; Δ H = x k J \frac{[H_{(a q) +}^{+} O H_{(a q)}^{-} H_{2} O_{(r) . Δ H = - 57.3 k J}] \times 2}{H_{2} C_{2} O_{4} + 2 O H^{-} \to 2 H_{2} O_{(r)} + C_{2} O_{4}^{2 -}, Δ H = x - 114.6! k J} B u t x - 114.6 = - 106 (g i v e n) \Rightarrow x = 8.6 k J m o l^{- 1}$

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When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous solution, 106 kJ of heat is liberated. Predict the enthalpy of ionization of the acid ?

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When 1 mole of oxalic acid is treated with excess of NaOH in dilute aqueous solution, 106 kJ of heat is liberated. Predict the enthalpy of ionization of the acid ?

The correct option is D 8.6 kJ mol−1 H2C2O4(aq)→2H+aq+C2O2−4(aq);ΔH=x kJ[H+(aq)+OH−(aq)H2O(r).ΔH=−57.3kJ]×2H2C2O4+2OH−→2H2O(r)+C2O2−4,ΔH=x−114.6!kJBut x−114.6=−106(given)⇒x=8.6kJ mol−1