The equilibrium in a saturated solution of Ca(OH)2 is represented above. In an experiment, a student places 5.0g of Ca(OH)2(s) i
nto 100.0mL of distilled water and stirs the mixture. How would the results be affected if the student repeats the experiment but this time places 5.0g of Ca(OH)2(s) into 100.0mL of 0.0010MNaOH(aq) instead of distilled
The answer is b) Less solid will disolve, because the larger value of [OH]- will cause the equilibrium position to lie farther to the left.
Explanation:
According to the reactions:
Ca(OH)2 = Ca2+ + 2OH-
NaOH = Na+ + OH-
The balance in the less solid solution will shift to the left, to keep the sodium hydroxide concentration in balance. This is produced by the common ionic effect in both reactions due to hydroxide (OH-) ions that are created due to sodium hydroxide NaOH, therefore, the solubility of calcium hydroxide Ca(OH)2 will be reduced. In conclusion we can say that the balance in the reaction will be more to the left.
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