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:

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:

There are a couple of key functional group spectra that you must memorize. A carbonyl group will cause a sharp dip at about 1700cm-1, and an alcohol group will cause a broad dip around 3400cm-1.

There are a couple of key functional group spectra that you must memorize. A carbonyl group will cause a sharp dip at about 1700cm-1, and an alcohol group will cause a broad dip around 3400cm-1.

There are a couple of key functional group spectra that you must memorize. A carbonyl group will cause a sharp dip at about 1700cm-1, and an alcohol group will cause a broad dip around 3400cm-1.

To solve this question you need to think about the chemical reaction occurring.

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.

If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.

L = mol/M = (0.5mol)/(2M) = 0.25L

| | Acetic acid | NaOH | Acetate | | | -------------- | --------- | --------- | -------- | | I | 1 mol | 0.5 mol | 0 mol | | C | -0.5 mol | -0.5 mol | +0.5 mol | | E | 0.5 mol | 0 mol | 0.5 mol |

To solve this question you need to think about the chemical reaction occurring.

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.

If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.

L = mol/M = (0.5mol)/(2M) = 0.25L

| | Acetic acid | NaOH | Acetate | | | -------------- | --------- | --------- | -------- | | I | 1 mol | 0.5 mol | 0 mol | | C | -0.5 mol | -0.5 mol | +0.5 mol | | E | 0.5 mol | 0 mol | 0.5 mol |

To solve this question you need to think about the chemical reaction occurring.

We can ignore water and sodium ions for the sake of this question. The reactants exist in a 1:1 ratio, so that for every mol of NaOH we add, we lose one mol of acetic acid and gain one mol of acetate. We can determine the moles of acetic acid by using M = mol/L, which gives us mol = ML = (2M) * (0.5L) = 1mol acetic acid. If we use the Hendersen Hasselbach equation we can see that the pH equals the pKa when the concentration of conjugate base (acetate) equals the concentration of acid.

If we have 1mol of acetic acid and add 0.5mol of NaOH, we will lose 0.5mol of acetic acid and gain 0.5mol of acetate. We will then be at a point where acetic acid equals acetate. This is summarized in the ICE table below. Now we know the moles of NaOH (0.5 moles) and the concentration (2M) so we can find the volume by doing M = mol/L.

L = mol/M = (0.5mol)/(2M) = 0.25L

| | Acetic acid | NaOH | Acetate | | | -------------- | --------- | --------- | -------- | | I | 1 mol | 0.5 mol | 0 mol | | C | -0.5 mol | -0.5 mol | +0.5 mol | | E | 0.5 mol | 0 mol | 0.5 mol |

H2CO3 will have two equivalence points and two half equivalence points, one set corresponding to each of its two protons.

After the first equivalence point, all of the acid is in the form HCO3-. When an acid is at a half equivalence point, the acid's concentration will be equal to the concentration of the conjugate base. For the second half equivalence point, the acid is in the HCO3- form, and the conjugate base is CO32-. As a result, at the second half equivalence point, the concentrations of HCO3- and CO32- will be equal.