Chemical Thermodynamics Question 337
Question: Heat of neutralization of a strong acid HA and a weaker acid HB with KOH are $ -13.7 $ and $ -12.7kcalmo{{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 $ HB=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 $
