# Question #7c173

Sep 16, 2015

You have the ${K}_{s p}$ expression. You have both $\left[A {g}^{+}\right]$ and $\left[N {O}_{2}^{-}\right]$. What you don't have is the final volume of solution; let's call this $2 V$ (because a volume, $V$, of silver nitrate, was mixed with an equal volume of sodium nitrite.

#### Explanation:

Moles of silver nitrate, $\left(A g N {O}_{3}\right)$: $0.040$ $m o l$ ${L}^{-} 1 \times V$.

Concentration of $\left[A g N {O}_{3}\right]$ when mixed: $0.040$ $m o l$ ${L}^{-} 1 \times V$/$2 V$ = $0.020$ $m o l$ ${L}^{-} 1$.

Concentration of $\left[A {g}^{+}\right]$ when mixed: $0.040$ $m o l$ ${L}^{-} 1 \times V$/$2 V$ = $0.020$ $m o l$ ${L}^{-} 1$. (Volume is 2V because equal volumes were mixed.)

Concentration of $\left[N {O}_{2}^{-}\right]$ when mixed: $0.030$ $m o l$ ${L}^{-} 1 \times V$/$2 V$ = $0.015$ $m o l$ ${L}^{-} 1$.

So we have $Q$, the ion product = $\left[A {g}^{+}\right] \left[N {O}_{2}^{-}\right]$ = $0.020 \times 0.015$ = ??

If $Q$ > ${K}_{s p}$, then precipitation of $A g N {O}_{2}$ will occur.