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A 100 ml mixture of Na(2)CO(3) and NaHCO...

A 100 ml mixture of `Na_(2)CO_(3)` and `NaHCO_(3)` is tittrated against 1 M HCl. If `v_(1)L` and `v_(2)L` are consumed when phenolphthalein and methyl orange are used as indicators respectively in two separate titrations, which of the following is true form molarities in the original solution.

A

molarity of `Na_(2)CO_(3)=20v_(1)`

B

molarity of `NaHCO_(3)=10(v_(2)-2v_(1))`

C

molarity of `Na_(2)CO_(3)=10(v_(2)+v_(1))`

D

molarity of `NaHCO_(3)=10(v_(2)-v_(1))`

Text Solution

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The correct Answer is:
To solve the problem, we need to analyze the titration of a mixture of sodium carbonate (Na₂CO₃) and sodium bicarbonate (NaHCO₃) with hydrochloric acid (HCl) using two different indicators: phenolphthalein and methyl orange. ### Step-by-step Solution: 1. **Understanding the Reaction with Phenolphthalein:** - When phenolphthalein is used, it indicates the endpoint of the reaction when all the Na₂CO₃ has reacted with HCl. - The reaction can be represented as: \[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] - For Na₂CO₃, 1 mole reacts with 2 moles of HCl. Thus, the milli-equivalents of Na₂CO₃ will equal the milli-equivalents of HCl at the endpoint. - If \( V_1 \) liters of HCl are consumed, then: \[ \text{milli-equivalents of Na}_2\text{CO}_3 = 2 \times V_1 \] 2. **Setting Up the Equation for Na₂CO₃:** - Let \( M_1 \) be the molarity of Na₂CO₃. The equation can be set up as: \[ M_1 \times \frac{100}{1000} = 2 \times V_1 \] - Rearranging gives: \[ M_1 = 20 \times V_1 \] 3. **Understanding the Reaction with Methyl Orange:** - Methyl orange indicates the endpoint when both Na₂CO₃ and NaHCO₃ have reacted with HCl. - The reactions are: - For Na₂CO₃: \[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] - For NaHCO₃: \[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] - The total milli-equivalents of Na₂CO₃ and NaHCO₃ will equal the milli-equivalents of HCl consumed, which is \( V_2 \) liters. - Thus: \[ 2 \times V_1 + V_{NaHCO_3} = V_2 \] - Rearranging gives: \[ V_{NaHCO_3} = V_2 - 2V_1 \] 4. **Setting Up the Equation for NaHCO₃:** - Let \( M_2 \) be the molarity of NaHCO₃. The equation can be set up as: \[ M_2 \times \frac{100}{1000} = V_{NaHCO_3} \] - Substituting \( V_{NaHCO_3} \): \[ M_2 \times \frac{100}{1000} = V_2 - 2V_1 \] - Rearranging gives: \[ M_2 = 10 \times (V_2 - 2V_1) \] ### Final Results: - The molarity of Na₂CO₃ is: \[ M_1 = 20 \times V_1 \] - The molarity of NaHCO₃ is: \[ M_2 = 10 \times (V_2 - 2V_1) \] ### Conclusion: The correct option from the given choices is: - **Molarity of NaHCO₃ is equal to \( 10 \times (V_2 - 2V_1) \)**.
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