Question #7c173

1 Answer
Sep 16, 2015

You have the #K_(sp)# expression. You have both #[Ag^+]# and #[NO_2^-]#. What you don't have is the final volume of solution; let's call this #2V# (because a volume, #V#, of silver nitrate, was mixed with an equal volume of sodium nitrite.

Explanation:

Moles of silver nitrate, #(AgNO_3)#: #0.040# #mol# #L^-1 xx V#.

Concentration of #[AgNO_3]# when mixed: #0.040# #mol# #L^-1 xx V#/#2V# = #0.020# #mol# #L^-1#.

Concentration of #[Ag^+]# when mixed: #0.040# #mol# #L^-1 xx V#/#2V# = #0.020# #mol# #L^-1#. (Volume is 2V because equal volumes were mixed.)

Concentration of #[NO_2^-]# when mixed: #0.030# #mol# #L^-1 xx V#/#2V# = #0.015# #mol# #L^-1#.

So we have #Q#, the ion product = #[Ag^+][NO_2^-]# = #0.020 xx 0.015# = ??

If #Q# > #K_(sp)#, then precipitation of #AgNO_2# will occur.