The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is
`Na_(2)CO_(3)(aq.)+2HCl(aq.)rarrNaCl(aq)+CO_(2)(g)H_(2)O(l)`
If you had the two solutions of the same concentration, you would have to use double volume of HCl to reach the equivalence point.
Indicators change their colours at the end point of the reaction and hence we are able to know the end points (equivalence points of reactions).
25 ml of `Na_(2)CO_(3)` solution requires 100 ml of 0.1 M HCl to reach end point with phenolphthalein indicator. Molarity of `HCO_(3^(-))` ions in the resulting solution is

Options

A. 0.008 M
B. 0.04M
C. 0.16 M
D. 0.08M

Explanation

To solve the question regarding the molarity of `HCO₃⁻` ions in the resulting solution after the reaction between sodium carbonate (`Na₂CO₃`) and dilute hydrochloric acid (`HCl`), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ \text{Na}_2\text{CO}_3 (aq) + 2 \text{HCl} (aq) \rightarrow 2 \text{NaCl} (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \]

### Step 2: Determine the moles of `HCl` used
Given that 100 mL of 0.1 M `HCl` is used:
\[ \text{Moles of } HCl = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} = 0.01 \, \text{mol} \]

### Step 3: Calculate the moles of `Na₂CO₃` that reacted
From the balanced equation, we see that 2 moles of `HCl` react with 1 mole of `Na₂CO₃`. Therefore, the moles of `Na₂CO₃` that reacted can be calculated as:
\[ \text{Moles of } Na_2CO_3 = \frac{\text{Moles of } HCl}{2} = \frac{0.01 \, \text{mol}}{2} = 0.005 \, \text{mol} \]

### Step 4: Calculate the moles of `HCO₃⁻` produced
During the reaction, `Na₂CO₃` reacts with `HCl` to form `NaHCO₃` (sodium bicarbonate) before further reacting to form `NaCl`. Therefore, for every mole of `Na₂CO₃`, one mole of `HCO₃⁻` is produced:
\[ \text{Moles of } HCO_3^- = \text{Moles of } Na_2CO_3 = 0.005 \, \text{mol} \]

### Step 5: Calculate the total volume of the resulting solution
The total volume of the resulting solution after the reaction is the sum of the volumes of the reactants:
\[ \text{Total Volume} = 25 \, \text{mL} + 100 \, \text{mL} = 125 \, \text{mL} = 0.125 \, \text{L} \]

### Step 6: Calculate the molarity of `HCO₃⁻`
Molarity is defined as moles of solute per liter of solution. Thus, the molarity of `HCO₃⁻` ions can be calculated as:
\[ \text{Molarity of } HCO_3^- = \frac{\text{Moles of } HCO_3^-}{\text{Total Volume in L}} = \frac{0.005 \, \text{mol}}{0.125 \, \text{L}} = 0.04 \, \text{M} \]

### Final Answer
The molarity of `HCO₃⁻` ions in the resulting solution is **0.04 M**.

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