If a g is the mass of `NaHC_(2)O_(4)` required to neutralize 100mL of 0.2M NaOH and b g that required to reduce 100mL of 0.02mL `KMnO_(4)` in acidic medium then:

Options

A. a=b
B. 2a=b
C. a=2b
D. none of these

Answer

To solve the problem, we need to calculate the mass of `NaHC₂O₄` required to neutralize 100 mL of 0.2 M NaOH and the mass required to reduce 100 mL of 0.02 M KMnO₄ in acidic medium. Let's break it down step by step.

### Step 1: Calculate the moles of NaOH
The first step is to calculate the number of moles of NaOH in 100 mL of a 0.2 M solution.

\[ \text{Moles of NaOH} = \text{Molarity} \times \text{Volume (in L)} \]
\[ = 0.2 \, \text{mol/L} \times 0.1 \, \text{L} = 0.02 \, \text{mol} \]

### Step 2: Determine the mass of NaHC₂O₄ required to neutralize NaOH
Since NaHC₂O₄ reacts with NaOH in a 1:1 ratio, the moles of NaHC₂O₄ required will also be 0.02 mol.

Next, we need to find the mass of NaHC₂O₄ required:

\[ \text{Mass of NaHC₂O₄} = \text{Moles} \times \text{Molar Mass} \]

Assuming the molar mass of NaHC₂O₄ is approximately 98 g/mol (you can verify the exact value):

\[ \text{Mass of NaHC₂O₄} = 0.02 \, \text{mol} \times 98 \, \text{g/mol} = 1.96 \, \text{g} \]

So, \( A = 1.96 \, \text{g} \).

### Step 3: Calculate the moles of KMnO₄
Now, we calculate the moles of KMnO₄ in 100 mL of a 0.02 M solution.

\[ \text{Moles of KMnO₄} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} \]

### Step 4: Determine the n-factor for KMnO₄ and NaHC₂O₄
In acidic medium, the n-factor for KMnO₄ (which reduces from MnO₄⁻ to Mn²⁺) is 5.

For NaHC₂O₄, the oxidation state of carbon changes from +3 to +4 when it is oxidized to CO₂. Thus, the n-factor for NaHC₂O₄ is 1.

### Step 5: Calculate the equivalents of KMnO₄
Using the n-factor, we can calculate the equivalents of KMnO₄:

\[ \text{Equivalents of KMnO₄} = \text{Moles} \times \text{n-factor} = 0.002 \, \text{mol} \times 5 = 0.01 \, \text{equivalents} \]

### Step 6: Calculate the mass of NaHC₂O₄ required to reduce KMnO₄
Since the equivalents of NaHC₂O₄ must equal the equivalents of KMnO₄, we can set up the equation:

\[ \text{Equivalents of NaHC₂O₄} = \text{Moles} \times \text{n-factor} \]
\[ 0.01 = \text{Moles of NaHC₂O₄} \times 1 \]

Thus, the moles of NaHC₂O₄ required is 0.01 mol.

Now, we can find the mass of NaHC₂O₄ needed:

\[ \text{Mass of NaHC₂O₄} = 0.01 \, \text{mol} \times 98 \, \text{g/mol} = 0.98 \, \text{g} \]

So, \( B = 0.98 \, \text{g} \).

### Step 7: Compare A and B
Now we have:
- \( A = 1.96 \, \text{g} \)
- \( B = 0.98 \, \text{g} \)

### Final Result
To find the ratio \( \frac{A}{B} \):
\[ \frac{A}{B} = \frac{1.96}{0.98} = 2 \]

### Conclusion
Thus, the answer is that \( A \) is twice \( B \).

---