GRE Subject Test: Chemistry : Analytical Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #51 : Analytical Chemistry

The  of  is . What is the molar solubility of  in water? 

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is: 

The  for the above equation is:

Due to the molar ratios of the species of , the concentration of  and  should be equal when dissolved:

Plugging  into the  equation gives:

Therefore, the solubility of   in water is 

Example Question #1 : Solubility And Precipitation

Calculate the molar solubility of  with  if enough  is added to raise the pH of the solution to pH 12.

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is: 

The  for the above equation is:

Due to the solubility of  in water, the concentration of  can be set to:

Therefore the concentration of  should be double that of :

The solution was adjusted to pH 12 using , therefore:

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.39.13 pm

Plugging these variables into the  equation gives:

Therefore, the  equation can be approximated to:

Therefore, the solubility of  in water is .

Example Question #51 : Analytical Chemistry

 has a  equal to . What would be the numerical expression used to determine the molar solubility (S) of ?

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is below: 

The Ksp expression for the above chemical equation is: 

Rearranging this equation to solve for solubility (S) gives:

Plugging the value for Ksp into the equation gives:

Example Question #51 : Analytical Chemistry

Calculate the molar solubility of  with a  if enough  is added to raise the pH of the solution to pH 10.

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is below: 

The  for the above equation is:

Due to the solubility of  in water, the concentration of  can be set to:

Therefore the concentration of  should be double that of :

The solution was adjusted to pH 10 using , therefore:

Using an ICE table to process the data:

Screen shot 2015 11 13 at 4.35.54 pm

Plugging  into the  equation gives:

Therefore, the  equation can be approximated to:

Therefore, the solubility of   in water is .

Example Question #52 : Analytical Chemistry

Calculate the molar solubility of  with  in a solution containing  .

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is below: 

The  for the above equation is:

Due to the solubility of  in water, the concentration of  can be set to be:

Therefore  concentration will be double:

Using an ICE table to process the data:

Screen shot 2015 11 09 at 9.29.39 pm

Plugging these variable into the  equation gives:

Therefore, the  equation can be approximated to:

Therefore, the solubility of   in water is .

Example Question #55 : Analytical Chemistry

Calculate the molar solubility of  with  in a solution containing  .

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is below: 

The  for the above equation is:

Due to the solubility of  in water, taking into account the concentration of  from  and  can be set to:

The  concentration will be double the concentration of the  coming from :

Using an ICE table to process the data:

Screen shot 2015 11 09 at 10.02.00 pm

Plugging these variables into the  equation gives:

Therefore, the  equation can be approximated to:

Therefore, the solubility of   in water is .

Example Question #53 : Analytical Chemistry

 has a  equal to . What would be the numerical expression used to determine the molar solubility (S) of ?

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is below: 

The Ksp expression for the above chemical equation is: 

Rearranging this equation to solve for solubility (S) gives:

Plugging the value for Ksp into the equation gives:

Example Question #51 : Analytical Chemistry

For the titration of  with  solution of , the end point occurs upon addition of  of the  solution. Determine the initial concentration of chloride that was present in the original solution.

Possible Answers:

Correct answer:

Explanation:

The equivalence point is when enough titrant has been added so that the number of moles of titrant equals the number of moles of analyte. We must first determine the number of moles of silver ions at the equivalence point.

At the equivalence point the following occurs in solution:

The number of moles of silver ions at equivalence point is calculated as follows:

Plugging these values into the equation gives:

Example Question #11 : Solubility And Solution Equilibrium

How many moles of fluoride ions are present in  of a completely saturated solution of lead (II) fluoride?

Possible Answers:

Correct answer:

Explanation:

Recall that the solubility product constant is given by the equation below, for a reaction in the following format.

Using our balanced reaction, we can find the solubility product equation for the dissociation of lead (II) fluoride.

For each molecule of lead (II) fluoride that dissolves, it produces one lead ion and two fluoride ions. We can conclude that the concentration of fluoride ions in solution will be twice the concentration of lead ions.

Use these variables in the solubility product equation, along with the given value from the question.

Now we can solve for the value of .

Remember that this value is equal to the concentration of lead ions in solution and half the concentration of fluoride ions in solution.

Example Question #53 : Analytical Chemistry

Calculate the molar solubility of  which has a .

Possible Answers:

Correct answer:

Explanation:

The equation for the dissolution of  in water is: 

The  for the above equation is:

Due to the solubility of  in water, the concentration of  can be set to:

Therefore the concentration of  should be double that of :

Plugging these variables into the  equation gives:

Now, we can solve for the x-variable by setting this value equal to the given .

Therefore, the solubility of  in water is .

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