question_answer

Two weak acids HX and HY have \[{{K}_{a}}\] values \[1.75\times {{10}^{-5}}\]and \[1.3\times {{10}^{-5}},\] respectively, at a certain temperature. An equimolar solution for mixture of two acids is partially neutralised by\[NaOH\]. How is the ratio of the contents of \[{{X}^{\bigcirc -}}\] and \[{{Y}^{\bigcirc -}}\] ions related to the values and molarity?

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 K