Question

Question: The two salts andare both far more soluble in water than either ${\mathbf{\text{BaSO}}}_{\mathbf{\text{4}}}{\mathbf{\text{(K}}}_{\mathbf{\text{SP}}}{\mathbf{\text{= 1.1×10}}}^{\mathbf{\text{- 10}}}{\mathbf{\text{)at25}}}^{\mathbf{0}}\mathbf{\text{C}}$at. Suppose ${\mathbf{\text{50mLof0.040MBaCl}}}_{\mathbf{\text{2}}}{\mathbf{\text{(aq)addedto50.0mLof0.20MAg}}}_{\mathbf{\text{2}}}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}\mathbf{\text{(aq)}}$of is added to of .Calculate the concentrations of, $\mathbf{}\mathbf{}{\mathbf{\text{SO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2 -}}}{\mathbf{\text{(aq)Cl}}}^{\mathbf{\text{-}}}{\mathbf{\text{(aq),Ba}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq)andAg}}}^{\mathbf{\text{+}}}\mathbf{\text{(aq)}}$ that remain in solution at equilibrium.

Short answer

Answer

Total concentration

$$\begin{gathered} {\text{[Ba}}^{\text{2 +}}\text{]=0.01M} \\ {\text{SO}}_{\text{4}}^{\text{2 -}}{\text{]=1.1×10}}^{\text{- 8}}\text{M} \\ {\text{[Ag}}^{\text{+}}{\text{]=8×10}}^{\text{- 9}}\text{M} \\ {\text{[Cl}}^{\text{-}}\text{]=0.02M} \end{gathered}$$

Step-by-step solution

The concentrations of ions before precipitation

When we mixed and, they dissociate into ions so the concentration of ions are:

$$\begin{gathered} {\text{[Ba}}^{\text{2 +}}\text{]=0.02M} \\ {\text{[Ag}}^{\text{+}}\text{]=0.01×2 = 0.02M} \\ {\text{[Cl}}^{\text{-}}\text{]=0.02×2 = 0.04M} \\ \text{[S}{{\text{O}}_{\text{4}}}^{\text{2 -}}\text{]=0.01M} \end{gathered}$$

In the reaction between Ba²⁺ and $\text{[S}{{\text{O}}_{\text{4}}}^{\text{2 -}}$, and$B{a}^{2+}$ is in excess.

The concentration of ${\text{[Ba}}^{\text{2 +}}$is:

${\text{[Ba}}^{\text{2 +}}\text{]=(0.02 - 0.01)M = 0.01M}$

The$K_{sp}$ is:${\text{K}}_{\text{SP}}{\text{= [Ba}}^{\text{2 +}}{\text{][SO\_4}}^{\text{2}}\text{]}$

The concentration of${\text{SO}}_{\text{4}}^{\text{2 -}}$ is:

$$\begin{gathered} \text{[S}{{\text{O}}_{\text{4}}}^{\text{2 -}}\text{]=}\frac{{\text{K}}_{\text{SP}}}{{\text{[Ba}}^{\text{2 +}}\text{]}} \\ \text{[S}{{\text{O}}_{\text{4}}}^{\text{2 -}}\text{]=}\frac{{\text{1.1×10}}^{\text{- 10}}}{\text{0.01}} \\ \text{[S}{{\text{O}}_{\text{4}}}^{\text{2 -}}{\text{]=1.1×10}}^{\text{- 8}}\text{M} \end{gathered}$$

Remaining equation at equilibrium

In the reaction between and, the is in excess. The concentration of is:

$\left[C{l}^{-}\right]=(0.04-0.02)M=0.02M$

The K_sp is:$K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

The concentration of ${\mathrm{Ag}}^{\text{+}}$is:

$$\begin{gathered} {\text{[Ag}}^{\text{+}}\text{]=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Cl}}^{\text{-}}\text{]}} \\ {\text{[Ag}}^{\text{+}}\text{]=}\frac{{\text{1.6×10}}^{\text{- 10}}}{\text{0.02}} \\ {\text{[Ag}}^{\text{+}}{\text{]=8×10}}^{\text{- 9}}\text{M} \end{gathered}$$

Total concentration:

$$\begin{gathered} {\text{[Ba}}^{\text{2 +}}\text{]=0.01M} \\ {\text{[SO}}_{\text{4}}^{\text{2 -}}{\text{]=1.1×10}}^{\text{- 8}}\text{M} \\ {\text{[Ag}}^{\text{+}}{\text{]=8×10}}^{\text{- 9}}\text{M} \\ {\text{[Cl}}^{\text{-}}\text{]=0.02M} \end{gathered}$$ .