A sample of $A g C l$ was treated with $5.00 m L$ of $1.5 M$ $N a_{2} C O_{3}$ solution to give $A g_{2} C O_{3}$ . The remaining solution contained $0.0026 g$ of $C l^{-}$ ions per litre. Calculate the solubility product AgCl. $K_{s} p$ $A g_{2} C O_{3}$ = $8.2 * 10^{- 12}$

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Solution

$1$ So,
$0.01$ moles of $N a_{2} C O_{3}$ produces $0.01$ moles of $A g_{2} C O_{3}$ in $5$ ml solution. I think the solubility of
$A g_{2} C O_{3}$ must be same as that of $N a_{2} C O_{3}$ .
Then,
ksp of $A g_{2} C O_{3}$ = $[2 \times S o l u b i l i t y] 2 \times [S o l u b i l i t y]$
or, Solubility $= 1.270334 \times 10^{- 4}$ = Concentration of $[A g +]$
$2.$ Now,
$[C l -] = 0.003 / 35.5 M = 8.45 \times 10^{- 5}$
Therefore, I think Ksp of $A g C l$ should be:
Ksp $= 1.270334 \times 10^{- 4} \times 8.45 \times 10^{- 5} = 1.073 \times 10^{- 8} .$

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