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A mixture contains 1.0 mole each of NaOH...

A mixture contains 1.0 mole each of NaOH,`Na_(2)``CO_(3)` and `NaHCO_(3)`. When half of mixture is titrated with HCl ,it required x mole of HCl in presence of phenolphthalein. In another experiment ,half of mixture required y mole of same HCl in presence of methyl orange. Find the value of (x+y).

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To solve the problem, we need to analyze the titration of a mixture containing 1 mole each of NaOH, Na2CO3, and NaHCO3 with HCl. We will consider two cases: one with phenolphthalein and the other with methyl orange. ### Step-by-Step Solution: **Step 1: Understanding the Mixture** - The mixture contains: - 1 mole of NaOH - 1 mole of Na2CO3 - 1 mole of NaHCO3 - Total moles in the mixture = 3 moles. **Step 2: Titration with Phenolphthalein (Case 1)** - When half of the mixture (1.5 moles total) is titrated with HCl in the presence of phenolphthalein, only NaOH and Na2CO3 will react. - The reaction is: - NaOH + HCl → NaCl + H2O (1:1 ratio) - Na2CO3 + 2HCl → 2NaCl + H2O + CO2 (2:1 ratio) - In half the mixture: - Moles of NaOH = 0.5 - Moles of Na2CO3 = 0.5 - Therefore, the equivalents of NaOH reacting with HCl: - For NaOH: 0.5 moles of NaOH will require 0.5 moles of HCl. - For Na2CO3: 0.5 moles of Na2CO3 will require 0.5 * 2 = 1 mole of HCl. - Total HCl required (x) in this case: \[ x = 0.5 + 1 = 1.5 \text{ moles} \] **Step 3: Titration with Methyl Orange (Case 2)** - In the second case, when half of the mixture is titrated with HCl in the presence of methyl orange, all three components will react: - NaOH reacts with HCl (1:1 ratio) - Na2CO3 reacts with HCl (2:1 ratio) - NaHCO3 reacts with HCl (1:1 ratio) - In half the mixture: - Moles of NaOH = 0.5 - Moles of Na2CO3 = 0.5 - Moles of NaHCO3 = 0.5 - Therefore, the equivalents of HCl required: - For NaOH: 0.5 moles of NaOH will require 0.5 moles of HCl. - For Na2CO3: 0.5 moles of Na2CO3 will require 0.5 * 2 = 1 mole of HCl. - For NaHCO3: 0.5 moles of NaHCO3 will require 0.5 moles of HCl. - Total HCl required (y) in this case: \[ y = 0.5 + 1 + 0.5 = 2 \text{ moles} \] **Step 4: Finding x + y** - Now we can find the total: \[ x + y = 1.5 + 2 = 3.5 \] ### Final Answer: The value of \( x + y \) is **3.5**. ---
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In the study of titration of NaOH and Na_(2)CO_(3) . NaOH and NaHCO_(3) , Na_(2)CO_(3) and NaHCO_(3) , phenophthalein and methyl orange are used as indicators. (a). When phenolphthalein is used as an indicator for the above mixture: (i). It indicates complete neutralisation of NaOH or KOH (ii). It indicates half neutralisation of Na_(2)CO_(3) because NaHCO_(3) is formed at the end point. (b). When methyl orange is used as an indicator for the above mixture (i). It indicates complete neutralisation of NaOH or KOH (ii). It indicates half neutralisation of Na_(2)CO_(3) because NaCl is formed at the end point. Q. 1 L solution of Na_(2)CO_(3) and NaOH was made in H_(2)O . 100 " mL of " this solution required 20 " mL of " 0.4 M HCl in the presence of phenolphthalein however, another 100 mL sample of the same solution required 25 " mL of " the same acid in the presence of methyl orange as indicator. What is the molar ratio of Na_(2)CO_(3) and NaOH in the original mixture.

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Given two mixtures: (I) NaOH and Na_(2)CO_(3) and (II) NaHCO_(3) and Na_(2)CO_(3) . 100 ml of mixture I required w and x ml of 1 M HCl is separate titrations using phenophthalein and methyl orange indicators while 100 ml of mixture II required y and z ml of same HCl solutioni in separate titrations using the same indicators.

50 " mL of " a solution containing 1 g each of Na_2CO_3 , NaHCO_3 and NaOH was titrated with N HCl. What will be the titre value if: (a). Only phenolphthalein is used as an indicator?

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1.9 gram mixture of Na_(2)CO_(3) and NaHCO_(3) is dissolved in water t prepare 100 ml solution. To neutralize this solution in pressure of phenophthaline (HPh) 10ml of 1N HCL solution is required. In another experiment in pressure of methyl orange (MeOH) 150mL . of 0.2N HNO_(3) solution is required ot neutralize the same solution find percentage composition of given mixture :-

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