The enthalpies of neutralization of a weak acid HA and a weak acid HB by NaOH are –6900 cal/equivale — Thermodynamics and Thermochemistry Chemistry Question
Question
The enthalpies of neutralization of a weak acid HA and a weak acid HB by NaOH are –6900 cal/equivalent and –2900 cal/equivalent, respectively. When one equivalent of NaOH is added to a solution containing one equivalent of HA and one equivalent of HB, the enthalpy change was –3900 cal. If the base is distributed between HA and HB in the ratio 1 : x, the value of ‘x’ is
Answer: 3
💡 Solution & Explanation
Let $f$ be the fraction of NaOH reacting with HA, and $(1-f)$ reacting with HB. Total enthalpy change $= f(-6900) + (1-f)(-2900) = -3900\text{ cal}$. Rearranging gives $-6900f - 2900 + 2900f = -3900 \implies -4000f = -1000 \implies f = 0.25$. So 25% reacts with HA and 75% reacts with HB. The ratio of HA to HB reacting is $0.25 : 0.75 = 1 : 3$. Thus $x = 3$.
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