A sample of `Na_2CO_3H_2O` weighing 0.62 g is added to 100 mL of 0.1 N `H_2SO_4`. Will the resulting solution be acidic, basic or neutral ?

Answer

To determine whether the resulting solution will be acidic, basic, or neutral after adding 0.62 g of `Na2CO3·H2O` to 100 mL of 0.1 N `H2SO4`, we can follow these steps:

### Step 1: Calculate the molar mass of `Na2CO3·H2O`
The molar mass of `Na2CO3·H2O` can be calculated as follows:
- Sodium (Na): 22.99 g/mol × 2 = 45.98 g/mol
- Carbon (C): 12.01 g/mol × 1 = 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
- Water (H2O): 18.02 g/mol × 1 = 18.02 g/mol

Adding these together:
\[ \text{Molar mass of } Na2CO3·H2O = 45.98 + 12.01 + 48.00 + 18.02 = 124.01 \text{ g/mol} \]

### Step 2: Calculate the equivalent weight of `Na2CO3·H2O`
The equivalent weight of `Na2CO3·H2O` is calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
Where \( n \) is the number of replaceable hydroxide ions (OH⁻) in the reaction. For `Na2CO3`, \( n = 2 \) (it can react with two protons from the acid).
\[ \text{Equivalent weight} = \frac{124.01 \text{ g/mol}}{2} = 62.005 \text{ g/equiv} \]

### Step 3: Calculate the equivalents of `Na2CO3·H2O`
Using the mass of `Na2CO3·H2O` provided:
\[ \text{Equivalents of } Na2CO3·H2O = \frac{\text{mass}}{\text{equivalent weight}} = \frac{0.62 \text{ g}}{62.005 \text{ g/equiv}} \approx 0.01 \text{ equiv} \]

### Step 4: Calculate the equivalents of `H2SO4`
The equivalents of sulfuric acid can be calculated using its normality and volume:
\[ \text{Equivalents of } H2SO4 = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Volume in L} = \frac{100 \text{ mL}}{1000} = 0.1 \text{ L} \]
\[ \text{Equivalents of } H2SO4 = 0.1 \text{ N} \times 0.1 \text{ L} = 0.01 \text{ equiv} \]

### Step 5: Compare the equivalents
Now we compare the equivalents of `Na2CO3·H2O` and `H2SO4`:
- Equivalents of `Na2CO3·H2O` = 0.01 equiv
- Equivalents of `H2SO4` = 0.01 equiv

Since both the acid and the base have the same number of equivalents, they will completely neutralize each other.

### Conclusion
The resulting solution will be neutral.

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