For each of the following compounds, decide whether the compound's solubility in aqueous solution changes with pH. If the solubi
1 answer:
Answer:
1. Ca(OH)₂ : Yes
Highest solubility = pH of 2
2. Cu Br : Yes
Highest solubility = pH of 4
3. Ca₃(PO₄)₂ : Yes
Highest solubility = pH of 2
Explanation:
From the common ion effect, the solubility of an ionic compound in a solution which already contains one of the ions in that compound will reduce. This is because, when an external stress is placed on a system in equilibrium, the equilibrium position will shift to remove the effect of that stress. Therefore, when more of the same ions are added to a solution already containing that ion, equilibrium will shift towards removal of the ion, thereby reducing solubility.
1. For Ca(OH)₂:
Ca(OH)₂ is a fairly soluble ionic compound whose dissociation equation is given below:
Ca(OH)₂ (s) ----> Ca²+ (aq) + 2 OH- (aq)
Increase in pH will result in addition of more OH- ions, therefore, its solubility will decrease. On the other hand, a decrease in pH will remove OH- ions, thereby increasing solubility. The pH of highest solubility is 2.
2. For CuBr:
CuBr is a slightly acidic salt as it is salt of a strong acid, HBr and a weak base Cu(OH)₂. Hydrolysis of the salt will result in an acidic medium:
2 CuBr (s) + 2 H₂O (l) ----> 2 HBr (aq) + Cu(OH)₂ (aq).
Thus, an increase in pH will result in an increase in the solubility of CuBr due to the removal of H+ ions. The highest solubility will be at pH of 4.
3) For Ca₃(PO₄)₂
Calcium phosphate is a salt of slightly basic salt as it is a salt of a weak acid, H₃PO₄ and stronger base Ca(OH)₂.
Hydrolysis of the salt will result in a basic medium:
Ca₃(PO₄)₂ (s) + 6 H₂O (l) ---> 3 Ca(OH)₂ (aq) + 2 H₃PO₄ (aq)
Therefore, its solubility increases with decrease in pH due to the removal of OH- ions. The highest solubility will be at pH of 2.
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Explanation:
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Answer: The rate of appearance of is
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
Rate in terms of disappearance of HBr = =
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The rate of appearance of ;
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Answer:
5-methylhex-1-yne//5-methylhex-2-yne//2-methylhex-3-yne
Explanation:
We have to start with the information on the <u>IR spectrum</u>. The signal at 3300 is due to a <u>C-H bend sp carbon</u> and the signal in 2200 is due to the <u>stretch carbon-carbon</u>. Therefore we will have an alkyne. Now if we have 2-methylhexane as the product of hydrogenation we have <u>several options to put the triple bond</u>. Between carbons 1 and 2 (5-methylhex-1-yne), between carbons 2 and 3 (5-methylhex-2-yne) and between carbons 3 and 4 (2-methylhex-3-yne). On carbon 5 we have a tertiary carbon therefore we dont have any other options.
See figure 1
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