Answer:
No precipitate will be formed.
Explanation:
Given that ;
150 mL of 0.15 M [tex]Na_2SO_4[/tex] is mixed with an equal volume of 0.050 M [tex]AgNO_3.[/tex]
Then;
[tex][Ag^+] = \frac{0.05*150}{(150+150)}[/tex]
[tex][Ag^+] = \frac{7.5}{(300.0)}[/tex]
[tex][Ag^+] = 0.025 \ M[/tex]
[tex][SO_4^{2-}] = \frac{0.15*150.0}{150.0+150.0}[/tex]
[tex][SO_4^{2-}] = \frac{22.5}{300.0}[/tex]
[tex][SO_4^{2-}] = 0.075 \ M[/tex]
For [tex]Ag_2SO_4[/tex] ; [tex][Ag^+]^2 [SO_4^{2-}][/tex]
[tex]=(0.025)^2 *(0.075)[/tex]
[tex]=4.6875*10^{-5}[/tex]
The Ksp of [tex]Ag_2SO_4[/tex] = [tex]1.5*10^{-5}[/tex] ;
Therefore since [tex]4.6875*10^{-5}[/tex] is < Ksp of [tex]Ag_2SO_4[/tex] ; Then no precipitate will be formed.