# If 2.5mL of 0.30M AgNO_3 is mixed with 7.5mL of 0.015M Na_2SO_4, should a precipitate of Ag_2SO_4 form? (Ksp = 1.2x10^-5)

Jul 3, 2016

Quite probably.

#### Explanation:

We need to work out the ion product for the following reaction, under the given conditions:

$2 A {g}^{+} + S {O}_{4}^{2 -} r i g h t \le f t h a r p \infty n s A {g}_{2} S {O}_{4} \left(s\right) \downarrow$

${Q}_{\text{the ion product}} = {\left[A {g}^{+}\right]}^{2} \left[S {O}_{4}^{2 -}\right]$

And now, we must determine the individual concentrations:

$\left[A {g}^{+}\right]$ $=$ $\frac{2.5 \times {10}^{- 3} \cdot L \times 0.30 \cdot m o l \cdot {L}^{-} 1}{\left(2.5 + 7.5\right) \times {10}^{-} 3 L}$ $=$ $7.50 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

$\left[S {O}_{4}^{2 -}\right]$ $=$ $\frac{7.5 \times {10}^{- 3} \cdot L \times 0.015 \cdot m o l \cdot {L}^{-} 1}{\left(2.5 + 7.5\right) \times {10}^{-} 3 L}$ $=$ $1.13 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

${Q}_{\text{the ion product}} = {\left(7.50 \times {10}^{-} 2\right)}^{2} \left(1.13 \times {10}^{-} 2\right) = 6.4 \times {10}^{-} 5$.

Since ${Q}_{\text{the ion product"=6.4xx10^-5>K_"sp}} = 1.2 \times {10}^{-} 5$, precipitation of silver sulfate should occur until equilibrium is satisfied, and $Q = {K}_{\text{sp}}$

Please don't trust my arithmetic. The approach I took (I think) was sound; we had to add volumes and recalculate concentrations.

Temperature was not referred to in this question; we would assume room temperature for ${K}_{\text{sp}}$. At higher temperatures, how would you expect ${K}_{\text{sp}}$ to evolve? Would solubility of silver sulfate increase or decrease?