- The change in color from blue to pink of the cobalt complexes here has been the basis of cobalt chloride indicator papers for the detection of the presence of water. It is also used in self-indicating silica gel desiccant granules.
- Pink cobalt species + chloride ions ⇌ Blue cobalt species + water molecules
<u>Explanation</u>:
- The adjustment in color from blue to the pink of the cobalt complexes here has been the premise of cobalt chloride indicator papers for the detection of the presence of water. It is likewise utilized in self-demonstrating silica gel desiccant granules.
Pink cobalt species + chloride particles ⇌ Blue cobalt species + water molecules
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The response of [Co(H2O)6]2+(aq) + 4Cl–(aq) → [CoCl4]2–(aq) + 6H2O(l) is endothermic. In this manner, as per Le Chatelier's rule, when the temperature is raised, the situation of the balance will move to one side, shaping a greater amount of the blue complex particle at the expense of the pink species.
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Including concentrated hydrochloric raises the chloride particle fixation, making the equilibrium move to one side, as per Le Chatelier. Including water brings down the chloride particle fixation, moving the equilibrium the other way.
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As an extension, it is conceivable to show that it is the Cl–particles in the hydrochloric acid that move the balance by including a spatula of sodium chloride rather than the pink arrangement. This delivers a bluer color, however, this may take some time because the salt is delayed to dissolve.
KOH is a strong base and HBr is a strong acid and completely dissociates.
The balanced equation for the reaction is;
KOH + HBr ---> KBr + H₂O
Stoichiometry of acid to base is 1:1
The number of KOH moles reacted - 0.50 M / 1000 mL/L x 48.0 mL = 0.024 mol
number of HBr moles reacted - 0.25 M/ 1000 mL/L x 96.0 mL = 0.024 mol
the number of H⁺ ions are equal to number of OH⁻ ions.
Then the solution is neutral.
pH of neutral solutions at 25 °C is 7.
Therefore pH is 7
Oxidation is when a substance gains oxygen molecules. For example when hydrogen reacts with oxygen it forms H₂O. The H₂ has been oxidised.
Answer:
The ΔG° is 29 kJ and the reaction is favored towards reactant.
Explanation:
Based on the given information, the ΔH°rxn or enthalpy change is 41.2 kJ, the ΔS°rxn or change in entropy is 42.1 J/K or 42.1 * 10⁻³ kJ/K. The temperature given is 289 K. Now the Gibbs Free energy change can be calculated by using the formula,
ΔG° = ΔH°rxn - TΔS°rxn
= 41.2 kJ - 289 K × 42.1 × 10⁻³ kJ/K
= 41.2 kJ - 12.2 kJ
= 29 kJ
As ΔG° of the reaction is positive, therefore, the reaction is favored towards reactant.
