SATHEE: Chemical Thermodynamics Question 337

Chemical Thermodynamics Question 337

Question: Heat of neutralization of a strong acid HA and a weaker acid HB with KOH are $- 13.7$ and $- 12.7 k c a l m o l^{- 1}$ . When 1 mole of KOH was added to a mixture containing 1 mole each of HA and HB, the heat change was $- 13.5$ kcal. In what ratio is the base distributed between HA and HB.

Options:

A) 3 : 1

B) 1 : 3

C) 4 : 1

D) 1 : 4

Show Answer

Answer:

Correct Answer: C

Solution:

Let x mole of KOH be neutralized by the strong acid HA. Then, moles neutralized by $H B = 1 - x$

Hence, $- 13.7 \times x + (- 12.7) \times (1 - x) = - 13.5$

$\Rightarrow x = 0.8; \frac{x}{1 - x} = \frac{0.8}{0.2} = \frac{4}{1} 4 : 1$

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Chemical Thermodynamics Question 337

Question: Heat of neutralization of a strong acid HA and a weaker acid HB with KOH are −13.7 and −12.7kcalmol−1 . When 1 mole of KOH was added to a mixture containing 1 mole each of HA and HB, the heat change was −13.5 kcal. In what ratio is the base distributed between HA and HB.

Options:

Answer:

Solution:

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