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The enthalpy of neutralization of a weak acid HA with NaOH is -6900 cal/equivalent, and that of weak acid HB with NaOH is -2900 cal/equivalent. When one equivalent of NaOH is added to a solution containing one equivalent of HA and one equivalent of HB, the observed enthalpy change is -3900 cal. If the NaOH gets distributed between HA and HB in the molar ratio X: Y, find the value of X + Y.

1. 7
2. 5
3. 3
4. 10

Correct answer: 7

Solution

Let fraction of base reacting with HA be X/(X+Y) and with HB be Y/(X+Y). Then total enthalpy = -6900*(X/(X+Y)) + (-2900)*(Y/(X+Y)) = -3900. Solving gives X:Y = 1:3, so if X=1 and Y=3, X+Y = 4. But if expressed as smallest integers times a common denominator yielding a nicer answer: X=1, Y=3 gives X+Y=4. Re-check with X+Y=4 as ratio and simplest integers: X=1,Y=3 -> X+Y=4.

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