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`        Please answer question number 10.plzzzzzZzzzzzzzzzzzzzzzzzzzzzz`
3 months ago

```							In the titration of a mixture of NaOH and Na2CO3 with a solution of HCl the following reaction occurs: 1) NaOH + HCl → NaCl + H2O This is a reaction of a strong base with a strong acid, at the equivalence point, there is a sharp increase in the pH of the solution - that is, a change in color in the phenolphthalein and methyl orange. 2) Na2CO3 + HCl → NaНCO3 + NaCl In reaction 2, the pH at the equivalence point of about 8.5 - this corresponds to the pH color change of phenolphthalein. 	NaНCO3 + HCl → NaCl + H2CO3 H2CO3 → H2O + CO2↑ In reaction 3  carbonic acid is formed, the pH at the equivalence point of about 4.5 - this corresponds to the pH change in color of methyl orange. With phenolphthalein, titration reactions occur 1 and 2.When methyl orange titration reactions take place 1 and 3. Thus, the content of Na2CO3 in the solution can be calculated using the reaction 3. Volume HCl spent for this reaction is equal to the difference in volumes HCl spent on titration using phenolphthalein and methyl orange. VHCl = 72 - 50 = 22 ml = 0,022 l The number of moles of Na2CO3 in 25 ml of the solution is equal to n Na2CO3 = 0.022 * 0.2 = 0.0044 mol Na2CO3 molar mass is 106 g / mol. Then the weight of Na2CO3 is 0.0044 * 106 = 0.466 g On titration of NaOH in 25 ml of spent 50 - 22 = 28ml = 0.028 liters of a solution of HCl The number of moles of NaOH in 25 ml of the solution is equal to n = 0,028 * 0.2 = 0.0056 molThe molar mass of NaOH is 40 g / mol.Then the weight of NaOH is 0.0056 * 40 = 0.224 g The weight ratio of NaOH and Na2CO3 in solution is equal to 0.224 / 0.466 = 0.480
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3 months ago
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