We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.

We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.

Two answer choices fall in this range: 4.2 and 4.7.

1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.

Recall that pKa is defined as follows:

Here, is the acid dissociation constant. is a measure of the equilibrium strength of an acid and is unique for each acid. The higher the value of , the stronger the acid; however, a particular acid’s value, and subsequently its strength, can never be changed. The only way you can change the of an acid is by changing the identity of the acid itself. This means that the pKa value of an acid is also always constant; therefore, you cannot decrease an acid’s pKa.

Using the definition of pKa, we can see that the pKa of an acid increases as you decrease the acid dissociation constant (). A strong acid will have a high and a low pKa.

The pKa of an acid can never be altered; therefore, changing the concentration of the acid will not alter the pKa of the acid. It might change the amount of hydrogen ions produced and alter the pH; however, the pKa of the acid will stay constant.

We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.

We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.

Two answer choices fall in this range: 4.2 and 4.7.

1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.

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:

We know that pKa is equal to –log(Ka). Thus, pKa of acetic acid is –log(1.8 * 10–5). This is not an easy problem to solve in your head, but there is a trick.

We know that 1 * 10–4 > 1.8 * 10–5 > 1 * 10–5, and we know that –log(1 * 10–4) = 4 and –log(1 * 10–5) = 5. Now we can conclude that our pKa is somewhere between 4 and 5.

Two answer choices fall in this range: 4.2 and 4.7.

1.8 * 10–5 is closer to 1 * 10–5 than it is to 1 * 10–4, so we can pick the answer closer to 5 than to 4 : 4.7.

Recall that pKa is defined as follows:

Here, is the acid dissociation constant. is a measure of the equilibrium strength of an acid and is unique for each acid. The higher the value of , the stronger the acid; however, a particular acid’s value, and subsequently its strength, can never be changed. The only way you can change the of an acid is by changing the identity of the acid itself. This means that the pKa value of an acid is also always constant; therefore, you cannot decrease an acid’s pKa.

Using the definition of pKa, we can see that the pKa of an acid increases as you decrease the acid dissociation constant (). A strong acid will have a high and a low pKa.

The pKa of an acid can never be altered; therefore, changing the concentration of the acid will not alter the pKa of the acid. It might change the amount of hydrogen ions produced and alter the pH; however, the pKa of the acid will stay constant.

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:

Recall that pKa is defined as follows:

Here, is the acid dissociation constant. is a measure of the equilibrium strength of an acid and is unique for each acid. The higher the value of , the stronger the acid; however, a particular acid’s value, and subsequently its strength, can never be changed. The only way you can change the of an acid is by changing the identity of the acid itself. This means that the pKa value of an acid is also always constant; therefore, you cannot decrease an acid’s pKa.

Using the definition of pKa, we can see that the pKa of an acid increases as you decrease the acid dissociation constant (). A strong acid will have a high and a low pKa.

The pKa of an acid can never be altered; therefore, changing the concentration of the acid will not alter the pKa of the acid. It might change the amount of hydrogen ions produced and alter the pH; however, the pKa of the acid will stay constant.

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:

First convert concentration of OH- into concentration of H+. Remember that Kw is 1*10-14.

Then, convert concentration of H+ into pH.

Alternatively, you can use the hydroxide concentration to solve for pOH and convert to pH.