56.0 g KOH, 138.0 g `K_(2)CO_(3)` and 100.0 g `KHCO_(3)` is dissolved in water and the solution is made 1 L. 10 " mL of " this stock solution is titrated with 2.0 M HCl. Which of the following statements is/are correct?

Options

A. When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 10 mL
B. When phenolphthalein is used as an indicator from the very beginning the titre value of HCl will be 40 mL.
C. When methyl orange is used as an indicator from the very beginning, the titre value of HCl will be 20 mL.
D. When methyl orange is used as an indicator after the first end point the titre value of HCl will be 10 mL.

Explanation

To solve the problem step by step, we will first calculate the moles of each compound, then determine their concentrations in the stock solution, and finally calculate the volume of HCl required for titration using both phenolphthalein and methyl orange as indicators.

### Step 1: Calculate the moles of each compound

1. **Potassium Hydroxide (KOH)**:
- Mass = 56.0 g
- Molar mass of KOH = 56.0 g/mol
- Moles of KOH = Mass / Molar mass = 56.0 g / 56.0 g/mol = 1.0 mol

2. **Potassium Carbonate (K2CO3)**:
- Mass = 138.0 g
- Molar mass of K2CO3 = 138.0 g/mol
- Moles of K2CO3 = Mass / Molar mass = 138.0 g / 138.0 g/mol = 1.0 mol

3. **Potassium Bicarbonate (KHCO3)**:
- Mass = 100.0 g
- Molar mass of KHCO3 = 100.0 g/mol
- Moles of KHCO3 = Mass / Molar mass = 100.0 g / 100.0 g/mol = 1.0 mol

### Step 2: Calculate the concentration of each compound in the stock solution

Since the total solution volume is 1 L (1000 mL):

1. **Concentration of KOH**:
- Concentration = Moles / Volume = 1.0 mol / 1 L = 1.0 M

2. **Concentration of K2CO3**:
- Concentration = Moles / Volume = 1.0 mol / 1 L = 1.0 M

3. **Concentration of KHCO3**:
- Concentration = Moles / Volume = 1.0 mol / 1 L = 1.0 M

### Step 3: Titration with HCl using phenolphthalein

1. **Reactions**:
- KOH + HCl → KCl + H2O (n-factor = 1)
- K2CO3 + HCl → KHCO3 + KCl (n-factor = 1)

2. **Calculate equivalents for KOH and K2CO3 in 10 mL of stock solution**:
- For KOH:
\[ \text{Equivalence of KOH} = n \times C \times V = 1 \times 1 \, \text{M} \times \frac{10 \, \text{mL}}{1000} = 0.01 \, \text{eq} \]
- For K2CO3:
\[ \text{Equivalence of K2CO3} = 1 \times 1 \, \text{M} \times \frac{10 \, \text{mL}}{1000} = 0.01 \, \text{eq} \]

3. **Total equivalence of substrate**:
\[ \text{Total equivalence} = 0.01 + 0.01 = 0.02 \, \text{eq} \]

4. **Equivalence of HCl**:
\[ \text{Equivalence of HCl} = n \times C \times V \implies 0.02 = 1 \times 2 \, \text{M} \times \frac{V_1}{1000} \]
\[ V_1 = \frac{0.02 \times 1000}{2} = 10 \, \text{mL} \]

### Step 4: Titration with HCl using methyl orange

1. **Reactions**:
- KOH + HCl → KCl + H2O (n-factor = 1)
- K2CO3 + 2 HCl → 2 KCl + H2O + CO2 (n-factor = 2)
- KHCO3 + HCl → KCl + H2O + CO2 (n-factor = 1)

2. **Calculate equivalents for KOH, K2CO3, and KHCO3**:
- For KOH:
\[ \text{Equivalence of KOH} = 1 \times 1 \, \text{M} \times \frac{10 \, \text{mL}}{1000} = 0.01 \, \text{eq} \]
- For K2CO3:
\[ \text{Equivalence of K2CO3} = 2 \times 1 \, \text{M} \times \frac{10 \, \text{mL}}{1000} = 0.02 \, \text{eq} \]
- For KHCO3:
\[ \text{Equivalence of KHCO3} = 1 \times 1 \, \text{M} \times \frac{10 \, \text{mL}}{1000} = 0.01 \, \text{eq} \]

3. **Total equivalence of substrate**:
\[ \text{Total equivalence} = 0.01 + 0.02 + 0.01 = 0.04 \, \text{eq} \]

4. **Equivalence of HCl**:
\[ \text{Equivalence of HCl} = n \times C \times V \implies 0.04 = 1 \times 2 \, \text{M} \times \frac{V_2}{1000} \]
\[ V_2 = \frac{0.04 \times 1000}{2} = 20 \, \text{mL} \]

### Conclusion

- **Using phenolphthalein**: 10 mL of HCl is required.
- **Using methyl orange**: 20 mL of HCl is required.