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NaoH and Na(2)CO(3) are dissolved in 200...

NaoH and `Na_(2)CO_(3)` are dissolved in 200 ml aqeous solution. In the presence of phenolpthaleim indicator, 17.5 ml of 0.1 HCl are used to titrated this solution. Now methyl orange is added in the same solution titrated and requires 2.5 ml of the same HCl. Calculate the mass of NaOH & `NaCO_(3)`.

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To solve the problem, we need to determine the mass of NaOH and Na2CO3 in a 200 ml aqueous solution based on the titration results with HCl using two different indicators: phenolphthalein and methyl orange. ### Step-by-Step Solution: 1. **Understanding the Reaction**: - Sodium hydroxide (NaOH) is a strong base, and sodium carbonate (Na2CO3) is a weak base that can react with HCl. - The reactions are as follows: - For NaOH: \[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \] - For Na2CO3: \[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \] 2. **Setting Up the Equations**: - Let \( x \) be the moles of NaOH and \( y \) be the moles of Na2CO3 in the solution. - The total volume of the solution is 200 ml, and the titration with phenolphthalein indicates that both NaOH and Na2CO3 react with HCl. The total moles of NaOH and Na2CO3 reacting with HCl can be expressed as: \[ x + y = \frac{0.1 \times 17.5}{1000} \] \[ x + y = 0.00175 \text{ moles} \] 3. **Using Methyl Orange**: - Methyl orange only indicates the presence of strong acids and bases. It reacts with Na2CO3, which means: \[ 2y = \frac{0.1 \times 2.5}{1000} \] \[ 2y = 0.00025 \text{ moles} \implies y = 0.000125 \text{ moles} \] 4. **Substituting for \( y \)**: - Substitute \( y \) back into the first equation: \[ x + 0.000125 = 0.00175 \] \[ x = 0.00175 - 0.000125 = 0.001625 \text{ moles} \] 5. **Calculating Mass of NaOH**: - The molar mass of NaOH is approximately 40 g/mol. - Therefore, the mass of NaOH is: \[ \text{Mass of NaOH} = x \times \text{Molar mass of NaOH} = 0.001625 \times 40 = 0.065 \text{ g} = 65 \text{ mg} \] 6. **Calculating Mass of Na2CO3**: - The molar mass of Na2CO3 is approximately 106 g/mol. - Therefore, the mass of Na2CO3 is: \[ \text{Mass of Na2CO3} = y \times \text{Molar mass of Na2CO3} = 0.000125 \times 106 = 0.01325 \text{ g} = 13.25 \text{ mg} \] ### Final Results: - Mass of NaOH = 65 mg - Mass of Na2CO3 = 13.25 mg
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