HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with .

47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.

HNO2 → H+ + NO2–

Initial 0.25M 0 0

Equilibrium 0.25 – x x x

Now, we can fill in our equation: .

Since x is very small, we can ignore it in the denominator:

(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)

Solve for x, and you find . Looking at our table, we know that

Now we can solve for pH:

HNO2 is a weak acid; it will not fully dissociate, so we need to use the HA → H+ + A– reaction, with .

47.0g HNO2 is equal to 1mol. 1mol into 4L gives a concentration of 0.25M when the acid is first dissolved; however, we want the pH at equilibrium, not at the initial state. As the acid dissolves, we know \[HNO2\] will decrease to become ions, but we don't know by how much so we indicate the decrease as "x". As HNO2 dissolves by a factor of x, the ion concentrations will increase by x.

HNO2 → H+ + NO2–

Initial 0.25M 0 0

Equilibrium 0.25 – x x x

Now, we can fill in our equation: .

Since x is very small, we can ignore it in the denominator:

(they expect you to do this on the MCAT; you will never have to solve with x in the denominator on the exam!)

Solve for x, and you find . Looking at our table, we know that

Now we can solve for pH:

HCl is a strong acid; it will fully dissociate in water, meaning that the concentration of H+ is equal to the concentration of HCl (they are in a 1 : 1 ratio). 7.0g of HCl is equal to 0.2mol (MW of HCl is 35.3g/mol). 0.2mol HCl goes into 3L of water, resulting in a concentration of 0.067M, or .

Now we know that and pH=-log\[H+\]. Using our trick for -log, we can see that . Since and , we know our answer is between 1 and 2.

is closer to , so we can pick the answer closer to 1. i.e. 1.2.

HCl is a strong acid; it will fully dissociate in water, meaning that the concentration of H+ is equal to the concentration of HCl (they are in a 1 : 1 ratio). 7.0g of HCl is equal to 0.2mol (MW of HCl is 35.3g/mol). 0.2mol HCl goes into 3L of water, resulting in a concentration of 0.067M, or .

Now we know that and pH=-log\[H+\]. Using our trick for -log, we can see that . Since and , we know our answer is between 1 and 2.

is closer to , so we can pick the answer closer to 1. i.e. 1.2.

Using the pH of 2.0, we can find that , because .

Since the acid is strong (fully dissociates) and monoprotic (one H+ per molecule),.

Our solution has 1L of water, meaning that we have of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of . Simplifying this ratio gives .

Using the pH of 2.0, we can find that , because .

Since the acid is strong (fully dissociates) and monoprotic (one H+ per molecule),.

Our solution has 1L of water, meaning that we have of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of . Simplifying this ratio gives .

A buffer must contain either a weak base and its salt or a weak acid and its salt. A mixture of ammonia and ammonium chloride is an example of the first case, since ammonia, NH3, is a weak base and ammonium chloride, NH4Cl, contains its salt.

Though autoionization of water produces small amounts of H3O+ and OH-, each conjugate salts of H2O, they exist in such small amounts as to make any buffering effects negligible.

A buffer must contain either a weak base and its salt or a weak acid and its salt. A mixture of ammonia and ammonium chloride is an example of the first case, since ammonia, NH3, is a weak base and ammonium chloride, NH4Cl, contains its salt.

Though autoionization of water produces small amounts of H3O+ and OH-, each conjugate salts of H2O, they exist in such small amounts as to make any buffering effects negligible.

In a buffered solution, when the concentrations of acid and conjugate base are equal, we know the .

This is derived from the Henderson-Hasselbalch equation: .

When concentrations are equal, the log of is , and .

In our question, the pH is approximately one unit greater than the .

Because of this, we know that the log of the two concentrations must be equal to one.

The log of is , and therefore the conjugate base must be ten times greater than the acid; therefore the ratio of acid to base is approximately .

We can quickly narrow this question to two possible answers, since the buffered solution is more basic than the , and thus we would expect there to be a greater concentration of base than acid in the solution.

In a buffered solution, when the concentrations of acid and conjugate base are equal, we know the .

This is derived from the Henderson-Hasselbalch equation: .

When concentrations are equal, the log of is , and .

In our question, the pH is approximately one unit greater than the .

Because of this, we know that the log of the two concentrations must be equal to one.

The log of is , and therefore the conjugate base must be ten times greater than the acid; therefore the ratio of acid to base is approximately .

We can quickly narrow this question to two possible answers, since the buffered solution is more basic than the , and thus we would expect there to be a greater concentration of base than acid in the solution.