Stomach acid, a dilute solution of HCl in water, can be neutralized by reaction with sodium hydrogen carbonate.
`NaHCO_(3_((aq)))+HCl_((aq)) to NaCl_((eq))+H_(2)O_((l))+CO_(2_(g))`
How many milliliters of 0.125 M `NaHCO_(3)` solution are needed to neutralize 18.0 mL of 0.100 M HCI?

Options

A. `14.4 ` mL
B. 12.0 mL
C. 14.0 mL
D. 13.2 mL

Explanation

To solve the problem of how many milliliters of 0.125 M NaHCO₃ solution are needed to neutralize 18.0 mL of 0.100 M HCl, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction between sodium hydrogen carbonate (NaHCO₃) and hydrochloric acid (HCl) is:
\[ \text{NaHCO}_3 (aq) + \text{HCl} (aq) \rightarrow \text{NaCl} (aq) + \text{H}_2O (l) + \text{CO}_2 (g) \]

### Step 2: Determine the number of equivalents of HCl
To find the number of equivalents of HCl, we can use the formula:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \]

For HCl:
- Molarity (M) = 0.100 M
- Volume = 18.0 mL = 0.018 L

Calculating the equivalents of HCl:
\[ \text{Number of equivalents of HCl} = 0.100 \, \text{mol/L} \times 0.018 \, \text{L} = 0.0018 \, \text{mol} \]

### Step 3: Determine the number of equivalents of NaHCO₃ needed
From the balanced equation, we see that 1 mole of NaHCO₃ reacts with 1 mole of HCl. Therefore, the number of equivalents of NaHCO₃ needed is the same as that of HCl:
\[ \text{Number of equivalents of NaHCO}_3 = 0.0018 \, \text{mol} \]

### Step 4: Calculate the volume of NaHCO₃ solution required
Using the formula for equivalents again:
\[ \text{Number of equivalents} = \text{Molarity} \times \text{Volume (L)} \]

For NaHCO₃:
- Molarity (M) = 0.125 M
- Let the volume be \( V \) L.

Setting up the equation:
\[ 0.0018 \, \text{mol} = 0.125 \, \text{mol/L} \times V \]

Solving for \( V \):
\[ V = \frac{0.0018 \, \text{mol}}{0.125 \, \text{mol/L}} = 0.0144 \, \text{L} \]

### Step 5: Convert volume from liters to milliliters
To convert liters to milliliters, multiply by 1000:
\[ V = 0.0144 \, \text{L} \times 1000 = 14.4 \, \text{mL} \]

### Final Answer
The volume of 0.125 M NaHCO₃ solution needed to neutralize 18.0 mL of 0.100 M HCl is **14.4 mL**.

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