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2 years ago

The equilibrium for the indicator methyl orange is HC14H14SO3(red) + H2O C14H14SO3−(yellow) + H3O+ The reaction is exothermic.

Chemistry

1 answer:

muminat2 years ago

4 0

Answer:

The solution will turn red.

Explanation:

HC₁₄H₁₄SO₃ + H₂O ⇌ HC₁₄H₁₄SO₃⁻ +H₃O⁺

(red) (yellow)

Methyl orange is a weak acid in which the ionized and unionized forms are distinct colours and are in equilibrium with each other,

At about pH 3.4, the two the forms are present in equal amounts, and the indicator colour is orange.

If you add more acid, you are disturbing the equilibrium.

According to Le Châtelier's Principle, when you apply a stress to a system at equilibrium, it will respond in such a way as to relieve the stress.

The system will try to get rid of the added acid, so the position of equilibrium will move to the left.

More of the unionized molecules will form, so the solution will turn red.

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$\large \boxed{109.17 \, ^{\circ}\text{C}}$

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$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

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