A) \[\left[ \frac{\alpha }{1-\alpha } \right]=\frac{1.75}{1.3}\times \left[ \frac{\alpha '}{1-\alpha '} \right],\]\[0,\] where \[\alpha \] and \[\alpha \] are ionised fractions of the acids HX and HY respectively.
B) The ratio is unrelated to the \[{{K}_{a}}\] values.
C) The ratio is unrelated to the molarity.
D) The ratio is unrelated to the pH of the solution.
Correct Answer: A
Solution :
[a] Let \[\alpha \] and \[\alpha '\] are the degree of dissociation of HX and HY at same concentration. \[{{K}_{HX}}=\frac{[\alpha ]\,[\alpha +\alpha ']C}{[1-\alpha ]}\] \[{{K}_{HY}}=\frac{[\alpha ']\,[\alpha +\alpha ']C}{[1-\alpha ]}\] \[\therefore \,\,\,\frac{{{K}_{HX}}}{{{K}_{HY}}}=\left[ \frac{\alpha }{1-\alpha '} \right]\times \left[ \frac{1-\alpha '}{\alpha '} \right]\] \[\left( \frac{\alpha }{1-\alpha } \right)=\frac{1.75\times {{10}^{-5}}}{1.3\times {{10}^{-5}}}\times \left[ \frac{\alpha '}{1-\alpha '} \right]\] The ratio is unrelated to molarity, pH and KYou need to login to perform this action.
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