Given,\[CaS{{O}_{4}}\] is somewhat soluble in water.
I.When \[{{H}_{2}}S{{O}_{4}}\] is added to a solution of \[CaS{{O}_{4}}\] the solubility of the \[CaS{{O}_{4}}\] will be increased.
II.The addition of \[{{H}_{2}}S{{O}_{4}}\] will lower the pH of the solution.
A.Statement I is true, Statement II is true.
B.Statement I is true, Statement II is false.
C.Statement I is false, Statement II is true.
D.Statement I is false, Statement II is false.
Answer
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Hint: We know that the reaction of dilute sulphuric acid with calcium sulfate, zinc sulfate, lead sulfate and sodium sulfate is a double decomposition reaction. The solubility rule for sulphates states that almost all the sulphates are soluble.
Complete answer:
As we know that when dilute sulphuric acid reacts with calcium sulfate a double decomposition reaction occurs. According to the solubility rule for sulphates, lead sulphate is not soluble in water and thus, lead sulphate precipitates out. This makes clear that the given statement I is incorrect.
Whereas statement II states that by the addition of sulphuric acid the pH of solution will be lower pH of solution. pH is the measure of acidity and basicity of an aqueous solution. pH is used to measure the number of free hydrogen ions in the water and pOH is the measure of free number of hydroxide ions in water. Here as we know that will give ion on reaction which means \[{{H}^{+}}\] ions will be greater as compared to \[O{{H}^{-}}\] ion. For the condition the nature of solution will be towards acidic. Also, as we know pH of solution is given as Acidic then comes Neutral and then Basic. Thus, \[{{H}_{2}}S{{O}_{4}}\] will lower the pH of the solution.
Therefore, the correct answer is option C i.e., statement I is false, and statement II is true.
Note:
Remember that the dilute sulphuric acid produces white precipitate when added to the solution of calcium sulfate. When dilute sulphuric acid reacts with calcium sulfate a double decomposition reaction occurs. In the reaction, calcium sulfate and nitric acid are produced.
Complete answer:
As we know that when dilute sulphuric acid reacts with calcium sulfate a double decomposition reaction occurs. According to the solubility rule for sulphates, lead sulphate is not soluble in water and thus, lead sulphate precipitates out. This makes clear that the given statement I is incorrect.
Whereas statement II states that by the addition of sulphuric acid the pH of solution will be lower pH of solution. pH is the measure of acidity and basicity of an aqueous solution. pH is used to measure the number of free hydrogen ions in the water and pOH is the measure of free number of hydroxide ions in water. Here as we know that will give ion on reaction which means \[{{H}^{+}}\] ions will be greater as compared to \[O{{H}^{-}}\] ion. For the condition the nature of solution will be towards acidic. Also, as we know pH of solution is given as Acidic then comes Neutral and then Basic. Thus, \[{{H}_{2}}S{{O}_{4}}\] will lower the pH of the solution.
Therefore, the correct answer is option C i.e., statement I is false, and statement II is true.
Note:
Remember that the dilute sulphuric acid produces white precipitate when added to the solution of calcium sulfate. When dilute sulphuric acid reacts with calcium sulfate a double decomposition reaction occurs. In the reaction, calcium sulfate and nitric acid are produced.
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