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Step-by-Step Solution
Step 1: Write the Dissociation Equations
We have two sparingly soluble salts, calcium carbonate ($\text{CaCO}_3$) and calcium oxalate ($\text{CaC}_2\text{O}_4$). Their dissociation in water can be represented as:
$\text{CaCO}_3 \longrightarrow \text{Ca}^{2+} + \text{CO}_3^{2-}$
$\text{CaC}_2\text{O}_4 \longrightarrow \text{Ca}^{2+} + \text{C}_2\text{O}_4^{2-}$
Let the solubility of $\text{CaCO}_3$ in the mixture be $x \, \text{M}$ and that of $\text{CaC}_2\text{O}_4$ be $y \, \text{M}$. Then total $[\text{Ca}^{2+}]$ contributed by both salts becomes $(x + y)\,\text{M}.$
Step 2: Write the Solubility Product ($K_{\text{sp}}$) Expressions
For $\text{CaCO}_3$,
$K_{\text{sp}}(\text{CaCO}_3) = [\text{Ca}^{2+}][\text{CO}_3^{2-}] = (x + y)(x).$
Given $K_{\text{sp}}(\text{CaCO}_3) = 4.7 \times 10^{-9}.$
Hence,
$(x + y)\,x = 4.7 \times 10^{-9} \quad \dots (1)$
For $\text{CaC}_2\text{O}_4$,
$K_{\text{sp}}(\text{CaC}_2\text{O}_4) = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}] = (x + y)(y).$
Given $K_{\text{sp}}(\text{CaC}_2\text{O}_4) = 1.3 \times 10^{-9}.$
Hence,
$(x + y)\,y = 1.3 \times 10^{-9} \quad \dots (2)$
Step 3: Relate x and y Using the Ratio
Divide equation (1) by equation (2):
$\frac{(x + y)\,x}{(x + y)\,y} = \frac{4.7 \times 10^{-9}}{1.3 \times 10^{-9}}.$
Simplifying, $(x + y)$ cancels out, giving
$\frac{x}{y} = \frac{4.7}{1.3} \approx 3.6.$
Therefore, $x = 3.6\,y.$
Step 4: Substitute x in Equation (2)
From equation (2),
$(x + y)\,y = 1.3 \times 10^{-9}.$
Substitute $x = 3.6\,y,$ so
$(3.6\,y + y)\,y = 1.3 \times 10^{-9}.$
Hence,
$4.6\,y^2 = 1.3 \times 10^{-9}.$
Solve for $y$:
$y^2 = \frac{1.3 \times 10^{-9}}{4.6}.$
$y = \sqrt{\frac{1.3 \times 10^{-9}}{4.6}} \approx 1.68 \times 10^{-5}\,\text{M}.$
Step 5: Find x
Since $x = 3.6\,y,$
$x = 3.6 \times 1.68 \times 10^{-5} \approx 6.048 \times 10^{-5}\,\text{M}.$
Step 6: Calculate Total Calcium Ion Concentration
Total calcium ion concentration is
$[\text{Ca}^{2+}] = x + y = 6.048 \times 10^{-5} + 1.68 \times 10^{-5}.$
$[\text{Ca}^{2+}] \approx 7.746 \times 10^{-5}\,\text{M}.$
Step 7: Final Answer
Therefore, the concentration of $\text{Ca}^{2+}$ ions in water is
$7.746 \times 10^{-5}\,\text{M}.$
