# GRE Subject Test: Chemistry : Analytical Chemistry

## Example Questions

### Example Question #1 : Limiting Reagent

Given the unbalanced equation below, how many grams of carbon dioxide will be produced from one mole of glucose and three moles of oxygen?

Explanation:

The first step to solve will be to balance the chemical reaction:

We see that we see that for every one mole of glucose used, six moles of carbon dioxide will be made. Similarly, for every six moles of oxygen used, six moles of carbon dioxide will be formed. For the reaction to carry out to completion, however, there must exist six moles of oxygen for every one mole of glucose. In the problem's circumstances, one of these compounds becomes the limiting reactant, in this case it is oxygen.

We only have three moles of oxygen, but we would need six to react all the given glucose, making oxygen the limiting reagent. We need to find the carbon dioxide produced from the limited amount of oxygen present. Use the molar ratio between oxygen and carbon dioxide and the molar mass of carbon dioxide to solve.

### Example Question #1 : Gravimetric Analysis

Household vinegar contains the organic compound acetic acid with chemical formula, . If a  vinegar sample contains  of acetic acid, calculate the percent (mass/mass) of the  in the vinegar sample.

Density of vinegar is the following:

Explanation:

Convert the moles of  to grams:

To calculate the percentage of  in the vinegar we need to use the following formula:

Therefore,

### Example Question #6 : Solubility And Equilibrium

Consider the following balanced equation for the solubility of barium hydroxide in an aqueous solution.

What is the solubility of barium hydroxide?

Explanation:

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction.

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated  and , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

### Example Question #41 : Analytical Chemistry

Write the solubility product, , for .

Explanation:

The solubility product constant () is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for  can be written as:

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

The product, , can be considered a constant expression called the solubility product constant () and the equation can be written in the form:

### Example Question #42 : Analytical Chemistry

Write the solubility product, , for .

Explanation:

The solubility product constant () is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for  can be written as:

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

The product, , can be considered a constant expression called the solubility product constant () and the equation can be written in the form:

### Example Question #43 : Analytical Chemistry

Write the  expression for the following equilibria:

None of these

Explanation:

The solubility product constant () is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution.The equilibrium between undissolved  and its ions dissolved in solution is:

The equilibrium constant for  can be written as:

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

The product, , can be considered a constant expression called the solubility product constant () and the equation can be written in the form:

### Example Question #44 : Analytical Chemistry

Considering that  for the equilibrium reaction:

What would be the  concentration for a solution buffered at a pH of 8.5?

Explanation:

The pH of the solution is , therefore the pOH would be  considering the relationship:

Therefore  based on the equation

The solubility product expression for  is :

By inserting the knowns in to the expression, gives:

Rearrange this equation

### Example Question #6 : Analyzing Solids

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

Explanation:

The equation for the dissolution of  in water is below:

The Ksp for the above equation is:

Due to the solubility of  of in water, the concentration of  and - should be equal:

Plug  into the Ksp equation:

Therefore:

The solubility of  in water is

### Example Question #45 : Analytical Chemistry

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

Explanation:

The equation for the dissolution of  in water is below:

The Ksp for the above equation is:

Due to the solubility of  of in water, the concentration of  and  should be equal:

Plugging X into the Ksp equation gives:

Therefore:

The solubility of CuBr in water is

### Example Question #46 : Analytical Chemistry

Which salt is insoluble in water?

Explanation:

Solubility rules must be followed for substances in aqueous media. Below are some of the solubility rules and they must be followed in the order given. For example, rule 1 should have precedence over rule 2.

1. All alkali metal  and  compounds are soluble.

2. Nitrate , perchlorate , chlorate, , and acetate  salts are soluble.

3. Silver, lead, and mercury salts are insoluble.

Thus, we see that silver chloride is insoluble due to rule 3.