$(a)$ Bond angle in $PH_{4}^{+}$ is higher than that in $PH_{3}$. Why?
$(b)$ What is formed when $PH_{3}$ reacts with an acid?

(N/A) In $PH_{3}$,$P$ is $sp^{3}$ hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair.
As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion,the tetrahedral shape associated with $sp^{3}$ bonding is changed to pyramidal.
$PH_{3}$ combines with a proton $(H^+)$ to form $PH_{4}^{+}$ in which the lone pair is absent.
Due to the absence of a lone pair in $PH_{4}^{+}$,there is no lone pair-bond pair repulsion.
Hence,the bond angle in $PH_{4}^{+}$ is higher than the bond angle in $PH_{3}$.
$(b)$ When $PH_{3}$ reacts with an acid,it acts as a Lewis base due to the presence of a lone pair on the $P$ atom and forms phosphonium salts (e.g.,$PH_{3} + HCl \rightarrow PH_{4}Cl$).