Of `PH_(3)` and `H_(2)S` which is more acidic and why ?

Explanation

The strength of an acid depends upon the stability of the anion (i.e., conjugate base) it gives after release of a proton. Now the disssociation of `PH_(3), H_(2)S` and HCl in aqueous solution occurs as follows :
`{:(PH_(3)(g) + aq rarr H^(+) (aq) + H_(2)P^(-)(aq)" ,"" "K_(a) = 1.6 xx 10^(-29)),(H_(2)S (g) + aq rarr H^(+) (aq) + HS^(-) (aq)" ,"" "K_(a) = 1.3 xx 10^(-7)):}`
Since S (EN = 2.5) is more electronegative than P (EN = 2.1), therefore, S in `HS^(-)` can accommodate the negative charge more easily than P in `H_(2)P^(-)`. In other words, `HS^(-)` is more stable than `H_(2)P^(-)`, i.e., `H_(2)S` can release a proton more easily than `PH_(3)` and hence `H_(2)S` is a stronger acid than `PH_(3)`.