AP Chemistry : Solutions
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Example Questions
Example Question #7 : Precipitates And Calculations
The 
The dissociation of calcium hydroxide in aqueous solution is:
The 
Given a 
Now, we can use basic algebra to solve for 
Since we set 
Example Question #1 : Precipitates And Calculations
What type of reaction is also known as a precipitation reaction?
Double replacement
Single replacement
Combustion
Combination
Decomposition
Double replacement
Double replacement reactions can be further categorized as precipitation reactions since it is possible to make a precipitate (solid) from mixing two liquids. Combustion reactions involve using oxygen to burn another species, and the products are carbon dioxide and water. Combination reactions involve the synthesis of one molecule from two separate ones; decomposition is the opposite.
Example Question #1 : Solubility And Equilibrium
Calculate the molar solubility of AgBr in 0.050 M AgNO3 at room temperature. The Ksp of AgBr is 5.4 x 10-13.
1.08 M
2.16 M
1.08 x 10-11 M
1.57 x 10-12 M
2.16 x 10-11 M
1.08 x 10-11 M
Example Question #1 : Solubility And Equilibrium
Would the molar solubility of Cr(OH)3 increase or decrease as the pH is lowered (i.e. made more acidic)?
Increase
There is no change
Decrease
Can not be determined
Increase
Since Cr(OH)3 is a basic salt, decreasing the pH makes it more soluble.
Example Question #2 : Solubility And Equilibrium
Calculate the molar solubility of SrF2 in 0.023M NaF. The Ksp for SrF2 is 4.3 x 10-9.
3.2 x 10-3 M
3.2 x 10-5 M
8.1 x 10-6 M
5.2 x 10-4 M
1.6 x 10-3 M
8.1 x 10-6 M
Example Question #41 : Solutions
Calculate the molar solubility of Mn(OH)2 at pH 9.5. The Ksp for Mn(OH)2 is 1.6 x 10-13.
2.1 x 10-5 M
1.5 x 10-3 M
1.6 x 10-4 M
2.4 x 10-3 M
3.5 x 10-4 M
1.6 x 10-4 M
Example Question #42 : Solutions
Calculate the molar solubility of CaF2 (Ksp = 3.9 x 10-11) in a room temperature solution of 0.010 M Ca(C2H3O2)2.
1.6 x 10-2 M
3.2 x 10-2 M
3.1 x 10-5 M
4.2 x 10-4M
3.7 x 10-4 M
3.1 x 10-5 M
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?
1.075 g / mL * 0.475 L
First, convert to mL
1.075 g / mL * 475 mL = 510.6 g
Example Question #43 : Solutions
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?
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 #1 : 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
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.
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