CollegeBoard is also known to provide answer options that are meant to confuse you into choosing the wrong
answer instead of the correct one. But fear not! We will show you how to carefully read the question and make
sure that you are confident in the answer you choose!
Question 1
Answer: (A)
In this first question, we see that a base Mg(OH)2 is being titrated by an acid HNO3. In order to determine the
mass of the initial base, we need to perform a series of titration calculations. This is a great example of when
your labs come in handy, since the lab can help you visualize what is going on.
First, we need to determine the reaction that is occurring when Mg(OH)2 is reacted with HNO3. We first write
out the equation we know, which is that the base and acid react to create water and an aqueous solution of the
remaining ions:
Mg(OH)2 + HNO3 → H2O + Mg2+ + NO3–
We now need to balance the above reaction. Hydrogen is already balanced, so the next element we move to is
Oxygen. We see that Oxygen is unbalanced with five oxygen on the left and four oxygen on the right. We can
balance Oxygen by:
Mg(OH)2 + 2HNO3 → 2H2O + Mg2+ + 2NO3–
, This gives us 8 oxygen molecules on either side. This simple balancing act has now balanced the entire equation
with one Mg molecule, 8 O molecules, 4 H molecules and 2 N molecules on either side.
Now we return to the titration problem. Since we know that the acid is being used as the titration solution, we
know that HNO3 is in the buret. We also know that 20 mL is needed to get to the end point since:
22.17 mL – 2.17 mL = 20 mL
Now that we know the molar mass of the base (58.3 grams), the molarity of the acid solution, and the amount of
acid needed to titrate the bass and the balanced equation, we can perform a simple stoichiometric equation. Make
sure to balance all your units and note that it takes 2 moles of HNO3 for every one mole of Mg(OH)2!
Question 2
Answer: (E)
Geometry of a molecule or compound is an important topic covered on the AP Chemistry exam so it is important
that you know your tetrahedrals from your trigonal bipyramidials! An easy hint is that you can almost always
answer instead of the correct one. But fear not! We will show you how to carefully read the question and make
sure that you are confident in the answer you choose!
Question 1
Answer: (A)
In this first question, we see that a base Mg(OH)2 is being titrated by an acid HNO3. In order to determine the
mass of the initial base, we need to perform a series of titration calculations. This is a great example of when
your labs come in handy, since the lab can help you visualize what is going on.
First, we need to determine the reaction that is occurring when Mg(OH)2 is reacted with HNO3. We first write
out the equation we know, which is that the base and acid react to create water and an aqueous solution of the
remaining ions:
Mg(OH)2 + HNO3 → H2O + Mg2+ + NO3–
We now need to balance the above reaction. Hydrogen is already balanced, so the next element we move to is
Oxygen. We see that Oxygen is unbalanced with five oxygen on the left and four oxygen on the right. We can
balance Oxygen by:
Mg(OH)2 + 2HNO3 → 2H2O + Mg2+ + 2NO3–
, This gives us 8 oxygen molecules on either side. This simple balancing act has now balanced the entire equation
with one Mg molecule, 8 O molecules, 4 H molecules and 2 N molecules on either side.
Now we return to the titration problem. Since we know that the acid is being used as the titration solution, we
know that HNO3 is in the buret. We also know that 20 mL is needed to get to the end point since:
22.17 mL – 2.17 mL = 20 mL
Now that we know the molar mass of the base (58.3 grams), the molarity of the acid solution, and the amount of
acid needed to titrate the bass and the balanced equation, we can perform a simple stoichiometric equation. Make
sure to balance all your units and note that it takes 2 moles of HNO3 for every one mole of Mg(OH)2!
Question 2
Answer: (E)
Geometry of a molecule or compound is an important topic covered on the AP Chemistry exam so it is important
that you know your tetrahedrals from your trigonal bipyramidials! An easy hint is that you can almost always
