Two weak acids `HX` and `HY` have `K_(a)` values `1.75 xx 10^(-5)` and `1.3 xx 10^(-5)`, respectively, at a certain temperature. An equimolar solution of mixture of two acids is parially neutralised by `NaOH`. How is the ratio of the contents of `X^(Theta)` and `Y^(Theta)` ions related to the `K_(a)` values and molarity?

Options

A. `[(alpha)/(1-alpha)] = (1.75)/(1.3) xx [(alpha')/(1-alpha')],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.

Answer

A

Explanation

Let `alpha` and `alpha'` are the degree of ionisation of `HX` and `HY` at same concentration.
`{:(HX,hArr,H^(o+),+,X^(Theta)),(1-alpha,,(alpha+alpha'),,alpha),(HY,hArr,H^(o+),+,Y^(Theta)),((1-alpha'),,(alpha+alpha'),,alpha'):}`
`K_(HY) = ([alpha][alpha+alpha']C)/([1-alpha])`
`K_(HY = ([alpha'][alpha+alpha']C)/([1-alpha'])`
`:. (K_(HX))/(K_(HY)) = [(alpha)/(1-alpha)] xx [(1-alpha')/(alpha')]`
`((alpha)/(1-alpha)) = (1.75 xx 10^(-5))/(1.3 xx 10^(-5)) xx [(alpha')/(1-alpha')]`
The ratio is unrelated to malority, `pH` and `K_(a)`.