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The solubility product of BaSO(4)" at " ...

The solubility product of `BaSO_(4)" at " 25^(@) C " is " 1.0 xx 10^(-9)`. What would be the concentration of `H_(2)SO_(4)` necessary to precipitate `BaSO_(4)` from a solution of `0.01" M Ba"^(+2)` ions

A

`10^(-9)`

B

`10^(-8)`

C

`10^(-7)`

D

`10^(-6)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the concentration of \( H_2SO_4 \) necessary to precipitate \( BaSO_4 \) from a solution of \( 0.01 \, M \) \( Ba^{2+} \) ions, we can follow these steps: ### Step 1: Write the dissociation equation for \( BaSO_4 \) The dissociation of barium sulfate can be represented as: \[ BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq) \] ### Step 2: Write the expression for the solubility product \( K_{sp} \) The solubility product (\( K_{sp} \)) for \( BaSO_4 \) can be expressed as: \[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \] Given that \( K_{sp} = 1.0 \times 10^{-9} \). ### Step 3: Substitute the known concentration of \( Ba^{2+} \) We know the concentration of \( Ba^{2+} \) ions is \( 0.01 \, M \) (or \( 1.0 \times 10^{-2} \, M \)). Substituting this into the \( K_{sp} \) expression gives: \[ 1.0 \times 10^{-9} = (1.0 \times 10^{-2})[SO_4^{2-}] \] ### Step 4: Solve for the concentration of \( SO_4^{2-} \) Rearranging the equation to find the concentration of sulfate ions: \[ [SO_4^{2-}] = \frac{1.0 \times 10^{-9}}{1.0 \times 10^{-2}} = 1.0 \times 10^{-7} \, M \] ### Step 5: Relate the concentration of \( SO_4^{2-} \) to \( H_2SO_4 \) Since \( H_2SO_4 \) is a strong acid, it completely dissociates in solution: \[ H_2SO_4 \rightarrow 2H^+ + SO_4^{2-} \] This means that the concentration of sulfate ions \( [SO_4^{2-}] \) produced from \( H_2SO_4 \) is equal to the concentration of \( H_2SO_4 \) itself. Therefore, to achieve \( [SO_4^{2-}] = 1.0 \times 10^{-7} \, M \), the concentration of \( H_2SO_4 \) must also be: \[ [H_2SO_4] = 1.0 \times 10^{-7} \, M \] ### Conclusion The concentration of \( H_2SO_4 \) necessary to precipitate \( BaSO_4 \) from a solution of \( 0.01 \, M \) \( Ba^{2+} \) ions is: \[ \boxed{1.0 \times 10^{-7} \, M} \] ---
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