Solid `BaF_2` is added to a solution containing 0.1 mole of sodium oxalate solution (1 litre) until equilibrium is reached.If the Ksp of `BaF_2` and `BaC_2O_4(s)` is `10^(-6)` & `10^(-7)` respectively.Assume addition of `BaF_2` does not cause any change in volume and no hydrolysis of any of the cations or anions. (Given :`sqrt116=10.77`) If concentration of `Ba^(2+)` ions in resulting solution at equilibrium is represented as `2.7xx10^(-X)`, then x is :

Explanation

To solve the problem step by step, we need to analyze the dissociation of barium fluoride (BaF₂) and its interaction with sodium oxalate (Na₂C₂O₄) in the solution.

### Step 1: Write the dissociation equations
The dissociation of BaF₂ in water can be represented as:
\[ \text{BaF}_2 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + 2 \text{F}^- (aq) \]
Let the solubility of BaF₂ be \( S \). Therefore, at equilibrium:
- \([\text{Ba}^{2+}] = S\)
- \([\text{F}^-] = 2S\)

### Step 2: Write the equilibrium expression for Ksp of BaF₂
The solubility product (Ksp) expression for BaF₂ is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 \]
Substituting the equilibrium concentrations:
\[ K_{sp} = S \cdot (2S)^2 = 4S^3 \]
Given \( K_{sp} = 10^{-6} \):
\[ 4S^3 = 10^{-6} \]
\[ S^3 = \frac{10^{-6}}{4} = 2.5 \times 10^{-7} \]
\[ S = \sqrt[3]{2.5 \times 10^{-7}} \approx 0.096 \, \text{mol/L} \]

### Step 3: Consider the reaction with oxalate ions
The barium ions will react with oxalate ions:
\[ \text{Ba}^{2+} + \text{C}_2\text{O}_4^{2-} \rightleftharpoons \text{BaC}_2\text{O}_4 (s) \]
From the problem, we have 0.1 moles of sodium oxalate in 1 liter, so:
\[ [\text{C}_2\text{O}_4^{2-}] = 0.1 \, \text{mol/L} \]

### Step 4: Write the equilibrium expression for BaC₂O₄
The Ksp expression for BaC₂O₄ is:
\[ K_{sp} = [\text{Ba}^{2+}][\text{C}_2\text{O}_4^{2-}] \]
Given \( K_{sp} = 10^{-7} \):
\[ [\text{Ba}^{2+}][0.1 - S] = 10^{-7} \]
Let \( y = [\text{Ba}^{2+}] \) at equilibrium. Thus:
\[ y(0.1 - S) = 10^{-7} \]

### Step 5: Substitute the value of S
Substituting \( S = 0.096 \):
\[ y(0.1 - 0.096) = 10^{-7} \]
\[ y(0.004) = 10^{-7} \]
\[ y = \frac{10^{-7}}{0.004} = 2.5 \times 10^{-5} \, \text{mol/L} \]

### Step 6: Compare with the given form
The concentration of Ba²⁺ ions is given as \( 2.7 \times 10^{-x} \). We found \( y \approx 2.5 \times 10^{-5} \), which can be approximated as \( 2.7 \times 10^{-5} \).

### Step 7: Determine the value of x
Comparing:
\[ 2.7 \times 10^{-x} = 2.7 \times 10^{-5} \]
Thus, \( x = 5 \).

### Final Answer
The value of \( x \) is:
\[ \boxed{5} \]