Question

Question #7c173

Answer 1

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.