AP Chemistry : Other Solution Concepts

Example Questions

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Example Question #1 : Other Solution Concepts

A solution on NaCl has a denisty of 1.075 g/mL. If there are 0.475 L of solution present, what is the mass?

510.6 g
0.510 g
441.9 g
2.2 g
2.263 g
Explanation:

1.075 g / mL * 0.475 L

First, convert to mL

1.075 g / mL475 mL = 510.6 g

Example Question #2 : Other Solution Concepts

5L of 0.1M NaCl and 10L of 0.2M NaI are combined in a single vessel.

What is the final concentration of sodium ions in solution?

Explanation:

NaCl and NaI are both highly soluble; thus both solutions can be treated as containing separate ions of sodium, chloride, and iodide.The final concentration can be found by finding the total number of moles of sodium ions and the total volume from both solutions.

We can find the moles of sodium ions from each solution by multiplying the volume by the molar concentration.

The total moles of sodium ions is:

Divide this by the total solution volume to find the final concentration:

Example Question #3 : Other Solution Concepts

Given that the pKa of acetic acid is 4.76, what is the percentage of the protonated form of acetic acid in a solution where the pH is 6?

There is not enough information given

Explanation:

In order to solve this problem, we first must use the Henderson-Hasselbalch equation:

Since we are given the pH of the solution and the pKa of acetic acid, we are able to solve for the ratio of conjugate base to acid:

Now that we have the ratio of conjugate base to acid, we need to calculate the percentage of the acid, or protonated form, in solution. To do this, it's important to realize that for every 17.38 moles of conjugate base, there is 1 mol of acid. Therefore, the total amount of acetic acid + acetate is equal to 18.38.

Example Question #4 : Other Solution Concepts

Suppose that two containers,  and , contain equal amounts of water. If 5 moles of  is added to solution  and 5 moles of glucose is added to solution , which solution will experience a greater increase in boiling point?

Both solutions will exhibit the same change in boiling point

Solution , because  is able to dissociate into  and  ions, thus resulting in a greater amount of particles dissolved in solution

Neither solution will experience a change in boiling point

Solution , because glucose has a greater molar mass than

There is not enough information to answer the question

Solution , because  is able to dissociate into  and  ions, thus resulting in a greater amount of particles dissolved in solution

Explanation:

In the question stem, we are told that equal molar amounts of  and glucose are added to containers  and , respectively. The change in boiling point of water is a colligative property that is dependent on the number of dissolved solute particles, regardless of their identity. The addition of 5 moles of  will result in approximately 10 moles of dissolved solute, since each mol of  can dissociate into two ions, according to the following reaction:

Glucose, on the other hand, does not dissociate and simply remains as intact molecules. Thus, the addition of 5 moles of glucose to container  results in 5 moles of dissolved solute. Since solution  contains approximately twice as many dissolved solute particles as does solution , it will experience a greater increase in the boiling point of water.

Example Question #5 : Other Solution Concepts

Approximately what is the pH of a  solution of  at ?

There is insufficient information to answer the question

Explanation:

We are given the concentration of  in solution and asked to find the pH. To do this, we must make use of the following equation:

It is also important to realize that  is a strong base and will thus dissociate completey according to the following reaction:

Thus, for every one mol of  that reacts, an equal number of moles of  will be produced. And since there are   to begin with, then   will be produced.

Remember that this calculated value so far is the pOH, not the pH! To calculate the pH, it is vital to remember that  at . Thus,

Example Question #6 : Other Solution Concepts

What volume of water must be added to 750mL of 0.050M sodium chloride () in order to achieve a final concentration of 0.015M?

Explanation:

For a solution of known volume and concentration (molarity in this case), the volume needed to dilute the solution to a desired concentration may be found using the formula:

Where  and  are the initial and final concentrations, and  and  are the initial and final volumes. So, for 750mL (0.750L) of a 0.050M solution diluted to 0.015M:

Solving for :

Now that we know the total volume needed, we may find the volume that must be added by subtracting the initial volume () from the final volume ():

1.75L of water must be added to 750mL of 0.050M  in order to achieve a final concentration of 0.015M

Example Question #7 : Other Solution Concepts

What is the pH of a 0.025M solution of hydrochloric acid ()?

Explanation:

Since  is a strong acid, calculations should be carried out assuming that the compound dissociates completely:

and  are produced in a 1:1 ratio to total dissolved , so the concentration of  in solution is the same as the concentration of :

pH is related to the concentration of :

Example Question #71 : Solutions

What is the osmotic pressure of a 5.0M solution of  at ?

Explanation:

Osmotic pressure is represented by:

Where  Van’t hoff factor, ,  gas constant ,  temperature in . The Van’t hoff factor is a unitless number that represents the amount of ionic species that the compound  will dissociate in solution.  is part of a large group of molecules classified as hydrocarbons which normally do not dissociate at all in solution. Therefore, .

Plug in known values and solve.

Example Question #72 : Solutions

A solution was prepared by dissolving 22.0 grams of  in water to give a 110mL solution. What is the concentration in molarity of this solution?

Explanation:

In order to calculate the concentration, we must use molarity formula:

We must use the molecular weight of  to calculate the moles of solute:

Example Question #73 : Solutions

A solution was prepared by dissolving 40.0 grams of  in water to give a 50mL solution. What is the concentration in molarity of this solution?