How many grams of `NaHCO_(3)` are required to neutralise 1 mL of 0.0902 N vinegar ?

Options

A. `8.4 xx 10^(-3)g`
B. `1.5xx10^(-3)g`
C. `0.758xx10^(-3)g`
D. `1.07xx10^(-3)g`

Explanation

To determine how many grams of sodium bicarbonate (NaHCO₃) are required to neutralize 1 mL of 0.0902 N vinegar, we can follow these steps:

### Step 1: Understand the Neutralization Reaction
Sodium bicarbonate (NaHCO₃) acts as a weak base and will react with acetic acid (the main component of vinegar) in a neutralization reaction. The reaction can be represented as:
\[ \text{NaHCO}_3 + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate the Number of Gram Equivalents of Vinegar
The normality (N) of vinegar is given as 0.0902 N. Since we are dealing with 1 mL of vinegar, we can calculate the number of gram equivalents of the acetic acid in the vinegar:
\[ \text{Number of gram equivalents of vinegar} = N \times V = 0.0902 \, \text{N} \times \frac{1 \, \text{mL}}{1000 \, \text{mL/L}} = 0.0902 \times 0.001 = 0.0000902 \, \text{equivalents} \]

### Step 3: Set Up the Equation for Neutralization
According to the principle of neutralization, the number of gram equivalents of the base (NaHCO₃) must equal the number of gram equivalents of the acid (vinegar):
\[ \text{Number of gram equivalents of NaHCO}_3 = \text{Number of gram equivalents of vinegar} \]
Thus,
\[ \text{Number of gram equivalents of NaHCO}_3 = 0.0000902 \, \text{equivalents} \]

### Step 4: Calculate the Mass of Sodium Bicarbonate
To find the mass of NaHCO₃ required, we need to use its equivalent weight. The molecular weight of NaHCO₃ is approximately 84 g/mol. Since NaHCO₃ provides one equivalent of bicarbonate ion (HCO₃⁻), its equivalent weight is the same as its molecular weight:
\[ \text{Mass} = \text{Number of equivalents} \times \text{Equivalent weight} \]
\[ \text{Mass of NaHCO}_3 = 0.0000902 \, \text{equivalents} \times 84 \, \text{g/equivalent} = 0.0075688 \, \text{g} \]

### Step 5: Convert to Milligrams
To express this in milligrams:
\[ 0.0075688 \, \text{g} = 7.5688 \, \text{mg} \]

### Final Answer
Approximately **7.57 mg** of NaHCO₃ is required to neutralize 1 mL of 0.0902 N vinegar.

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