`40 mL 0.05 M` solution of sodium sesquicarbonate dehydrate `(Na_(2)CO_(3).NaHCO_(3).2H_(2)O)` is titrated against `0.05 M HCl` solution, `x mL` of acid is required to reach the phenolphthalein end point while mL of same acid were required when methyl organe indicator was used in a separate titration. Which of the following is (are) correct statements?
a.`y-x = 80 mL`
b. `y+x = 160 mL`
c. If the titration is started with phenolphthalein indicator and methyl orange is added at the end point, `2 x mL` of `HCl` would be required further to reach the end point
d. If the same volume of same solution is titrated against `0.10 M NaOH, x//2 mL` of base would be required

Explanation

To solve the problem step by step, we will analyze the titration of sodium sesquicarbonate dehydrate (Na₂CO₃·NaHCO₃·2H₂O) against hydrochloric acid (HCl) using two different indicators: phenolphthalein and methyl orange.

### Step 1: Calculate the moles of sodium sesquicarbonate dehydrate

Given:
- Volume of solution = 40 mL = 0.040 L
- Molarity of solution = 0.05 M

Using the formula for moles:
\[ \text{Moles} = \text{Molarity} \times \text{Volume} = 0.05 \, \text{mol/L} \times 0.040 \, \text{L} = 0.002 \, \text{mol} \]

### Step 2: Determine the moles of Na₂CO₃ and NaHCO₃

Since the formula of sodium sesquicarbonate dehydrate is Na₂CO₃·NaHCO₃·2H₂O, it contains:
- Moles of Na₂CO₃ = 0.002 mol
- Moles of NaHCO₃ = 0.002 mol

### Step 3: Titration with phenolphthalein

In the titration with phenolphthalein, only Na₂CO₃ reacts with HCl to form NaHCO₃. The reaction can be represented as:
\[ \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl} \]

The equivalence of Na₂CO₃ to HCl can be calculated as:
\[ \text{Equivalence of Na}_2\text{CO}_3 = \text{n-factor} \times \text{moles} = 1 \times 0.002 = 0.002 \, \text{eq} \]

For HCl:
\[ \text{Equivalence of HCl} = \text{Molarity} \times \text{Volume in L} = 0.05 \times \frac{x}{1000} \]

Setting the equivalences equal:
\[ 0.002 = 0.05 \times \frac{x}{1000} \]
\[ x = \frac{0.002 \times 1000}{0.05} = 40 \, \text{mL} \]

### Step 4: Titration with methyl orange

In the titration with methyl orange, both Na₂CO₃ and NaHCO₃ react with HCl:
1. Na₂CO₃ reacts to form NaHCO₃.
2. NaHCO₃ reacts to form H₂CO₃.

The total equivalence can be calculated as:
\[ \text{Equivalence of Na}_2\text{CO}_3 + \text{Equivalence of NaHCO}_3 = (2 \times 0.002) + (1 \times 0.002) = 0.006 \, \text{eq} \]

Setting this equal to HCl:
\[ 0.006 = 0.05 \times \frac{y}{1000} \]
\[ y = \frac{0.006 \times 1000}{0.05} = 120 \, \text{mL} \]

### Step 5: Analyze the options

Now we have:
- \( x = 40 \, \text{mL} \)
- \( y = 120 \, \text{mL} \)

1. **Option a:** \( y - x = 80 \, \text{mL} \) → **True**
2. **Option b:** \( y + x = 160 \, \text{mL} \) → **True**
3. **Option c:** If phenolphthalein is used first, then \( 2x \) mL of HCl would be required further to reach the endpoint → **True** (since \( y - x = 80 \, \text{mL} \) implies \( 2x = 80 \))
4. **Option d:** If titrated against 0.1 M NaOH, \( \frac{x}{2} \) mL of base would be required → **True** (since \( x = 40 \, \text{mL} \) implies \( \frac{40}{2} = 20 \, \text{mL} \))

### Conclusion

All options are correct.