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MCAT Physical : Acid-Base Chemistry

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Example Questions

Example Question #1 : Acid Base Equilibrium

Acids and bases can be described in three principal ways. The Arrhenius definition is the most restrictive. It limits acids and bases to species that donate protons and hydroxide ions in solution, respectively. Examples of such acids include HCl and HBr, while KOH and NaOH are examples of bases. When in aqueous solution, these acids proceed to an equilibrium state through a dissociation reaction.

$HA \leftrightharpoons H^{+} + A^{-}$ $\displaystyle HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$ $\displaystyle BOH \leftrightharpoons B^{+} + OH^{-}$

The Brønsted-Lowry definition of an acid is a more inclusive approach. All Arrhenius acids and bases are also Brønsted-Lowry acids and bases, but the converse is not true. Brønsted-Lowry acids still reach equilibrium through the same dissociation reaction as Arrhenius acids, but the acid character is defined by different parameters. The Brønsted-Lowry definition considers bases to be hydroxide donors, like the Arrhenius definition, but also includes conjugate bases such as the A^- in the above reaction. In the reverse reaction, A^- accepts the proton to regenerate HA. The Brønsted-Lowry definition thus defines bases as proton acceptors, and acids as proton donors.

Which of the following expressions most closely approximates the equilibrium constant of pure water?

Possible Answers:

$K_{w} = [}H^{+}}][OH^{-}]$ $\displaystyle K_{w} = [}H^{+}}][OH^{-}]$

$K_{w} = [}H^{+}}][H_{2}O]$ $\displaystyle K_{w} = [}H^{+}}][H_{2}O]$

$K_w=\frac{[H^+]}{[OH^-]}$ $\displaystyle K_w=\frac{[H^+]}{[OH^-]}$

$K_{w} = [H_{2}O][OH^{-}]$ $\displaystyle K_{w} = [H_{2}O][OH^{-}]$

$K_{w} = [}H^{+}}]$ $\displaystyle K_{w} = [}H^{+}}]$

Correct answer:

$K_{w} = [}H^{+}}][OH^{-}]$ $\displaystyle K_{w} = [}H^{+}}][OH^{-}]$

Explanation:

The autionization of water proceeds as follows:

$H_{2}O \leftrightharpoons H^{+} + OH^{-}$ $\displaystyle H_{2}O \leftrightharpoons H^{+} + OH^{-}$

The equilibrium constant is calculated by the following formula.

$K_{eq}=\frac{[product]}{[reactant]}$ $\displaystyle K_{eq}=\frac{[product]}{[reactant]}$

The equilibrium expression is $K_{w} = [}H^{+}}][OH^{-}]$ $\displaystyle K_{w} = [}H^{+}}][OH^{-}]$, with the pure water reactant excluded. Pure solids and liquids are not included in the equilibrium calculation.

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Example Question #28 : Mcat Physical Sciences

What is the hydroxide concentration in an aqueous solution with a pH of 2?

Possible Answers:

$1*10^{-4}M$ $\displaystyle 1*10^{-4}M$

$1*10^{-12}M$ $\displaystyle 1*10^{-12}M$

There are no hydroxide ions in the solution

$1*10^{-2}M$ $\displaystyle 1*10^{-2}M$

Correct answer:

$1*10^{-12}M$ $\displaystyle 1*10^{-12}M$

Explanation:

The hydroxide concentration can be determined by considering the autoionization of water in a solution. At 25^oC $\displaystyle 25^oC$, water has the equilibrium constant of $K_{w} = 1*10^{-14}$ $\displaystyle K_{w} = 1*10^{-14}$. This value is based on the autoionization of water.

$H_2O\rightleftharpoons H^++OH^-$ $\displaystyle H_2O\rightleftharpoons H^++OH^-$

$K_w=[H^+][OH^-]=1*10^{-14}$ $\displaystyle K_w=[H^+][OH^-]=1*10^{-14}$

Since the reaction is in terms of proton and hydroxide ion concentrations, we can set the expression equal to this value in order to determine the concentration of hydroxide ions at the given pH. We start by determining the concentration of protons in the solution by using the following equation:

$\displaystyle pH=-log[H^+]$

$[H^{+}] = 10^{-pH}$ $\displaystyle [H^{+}] = 10^{-pH}$

$[H^{+}] = 10^{-2} = 1*10^{-2}M$ $\displaystyle [H^{+}] = 10^{-2} = 1*10^{-2}M$

Now, we can solve for the hydroxide concentration.

$K_w=[H^+][OH^-]=1*10^{-14}$ $\displaystyle K_w=[H^+][OH^-]=1*10^{-14}$

$[OH^{-}][1*10^{-2}] = 1*10^{-14}$ $\displaystyle [OH^{-}][1*10^{-2}] = 1*10^{-14}$

$[OH^{-}] = 1*10^{-12}M$ $\displaystyle [OH^{-}] = 1*10^{-12}M$

This problem can also be solved by calculating the pOH of the solution and using this value to find the hydroxide ion concentration.

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Example Question #1 : Acid Base Chemistry

What is the pH of a solution which has a hydroxide ion concentration of 5 * 10^-4M?

Possible Answers:

$\small 10.70$ $\displaystyle \small 10.70$

$\small 5*10^{10}$ $\displaystyle \small 5*10^{10}$

$\small 2*10^{-11}$ $\displaystyle \small 2*10^{-11}$

$\small 3.30$ $\displaystyle \small 3.30$

$\small 13.30$ $\displaystyle \small 13.30$

Correct answer:

$\small 10.70$ $\displaystyle \small 10.70$

Explanation:

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

$\small [H^{+}]=\frac{1*10^{-14}}{[OH^{-}]} = \frac{1*10^{-14}}{5*10^{-4}}=2*10^{-11}$ $\displaystyle \small [H^{+}]=\frac{1*10^{-14}}{[OH^{-}]} = \frac{1*10^{-14}}{5*10^{-4}}=2*10^{-11}$

Then, convert concentration of H⁺ into pH.

$\small pH = -log[H^{+}]=-log(2*10^{-11})\approx10.70$ $\displaystyle \small pH = -log[H^{+}]=-log(2*10^{-11})\approx10.70$

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

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Example Question #1 : Acid Base Chemistry

Diffusion can be defined as the net transfer of molecules down a gradient of differing concentrations. This is a passive and spontaneous process and relies on the random movement of molecules and Brownian motion. Diffusion is an important biological process, especially in the respiratory system where oxygen diffuses from alveoli, the basic unit of lung mechanics, to red blood cells in the capillaries.

Capture

Figure 1 depicts this process, showing an alveoli separated from neighboring cells by a capillary with red blood cells. The partial pressures of oxygen and carbon dioxide are given. One such equation used in determining gas exchange is Fick's law, given by:

ΔV = (Area/Thickness) · D_gas · (P_{1 –}P₂)

Where ΔV is flow rate and area and thickness refer to the permeable membrane through which the gas passes, in this case, the wall of the avlveoli. P₁and P₂refer to the partial pressures upstream and downstream, respectively. Further, D_gas, the diffusion constant of the gas, is defined as:

D_gas= Solubility / (Molecular Weight)^(1/2)

Which following pair gives the correct equation, and change in pH, when carbon dioxide diffuses into blood?

Possible Answers:

CO₂ + H₂O H₂CO₃| pH increases

CO₂ + H₂O H₂CO₃| pH decreases

CO₂ + H₂O HCO₃^–| pH decreases

O₂ + CO₂+ H₂O H₂O₂ + CO₃^2–| pH decreases

Correct answer:

CO₂ + H₂O H₂CO₃| pH decreases

Explanation:

This is essentially a two-part question. The first asks you to generate the production of carbonic acid; and the second, requires you to infer that an acid will lower blood pH. Only one answer choice is both balanced correctly and gives an accurate change in pH. Although this knowledge is not required for the MCAT, your body constantly monitors blood pH to determine if concentrations of CO₂are too high, or alternatively, if O₂concentrations are too low.

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Example Question #1 : Acid Base Chemistry

As the value of K_a increases, __________

Possible Answers:

pKa decreases

all of these are true

the strength of the conjugate base decreases

the reaction HA → H⁺+ A^– favors the products

the strength of the acid increases

Correct answer:

all of these are true

Explanation:

K_a represents the equilibrium constant for the acid dissociation in water, HA → H⁺+ A^–, so it is a measure of the products divided by the reactants. As this value increases, the reaction favors the products (the ions) more, meaning that the acid dissociates more. The definition of a strong acid is one that fully dissociates in water, so as K_a increases the strength of the acid increases, and the strength of the conjugate base decreases. pK_a is defined as the –log(K_a), so as K_a increases pK_a decreases. All of these answer choices are correct.

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Example Question #2 : Reactions And Titrations

What is the pK_a of acetic acid? (K_a= 1.8 * 10^–5)

Possible Answers:

5.3

2.1

4.2

8.6

4.7

Correct answer:

4.7

Explanation:

We know that pK_a is equal to –log(K_a). Thus, pK_a 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 pK_a 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.

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Example Question #1 : Acid Base Equilibrium

HCN dissociates based on the following reaction.

$HCN + H_{2}O \rightarrow CN^{-}+ H_{3}O^{+}$ $\displaystyle HCN + H_{2}O \rightarrow CN^{-}+ H_{3}O^{+}$

The K_a for hydrogen cyanide is $\small 6.2*10^{-10}$ $\displaystyle \small 6.2*10^{-10}$.

What is the K_bfor CN^-?

Possible Answers:

$\small 6.2*10^{4}$ $\displaystyle \small 6.2*10^{4}$

$\small 1.6*10^{-5}$ $\displaystyle \small 1.6*10^{-5}$

$\small 1.3*10^{-4}$ $\displaystyle \small 1.3*10^{-4}$

$\small 6.2*10^{-24}$ $\displaystyle \small 6.2*10^{-24}$

Correct answer:

$\small 1.6*10^{-5}$ $\displaystyle \small 1.6*10^{-5}$

Explanation:

Remember that the K_a for the acid and the K_b for the conjugate base, when multiplied will equal the autoionization of water constant (K_w).

$K_{a} K_{b} = K_{w}=\small 1*10^{-14}$ $\displaystyle K_{a} K_{b} = K_{w}=\small 1*10^{-14}$

$\small K_{b} = \frac{K_w}{K_a}=\frac{1*10^{-14}}{6.2*10^{-10}}$ $\displaystyle \small K_{b} = \frac{K_w}{K_a}=\frac{1*10^{-14}}{6.2*10^{-10}}$

$\small K_{b}= 1.6*10^{-5}$ $\displaystyle \small K_{b}= 1.6*10^{-5}$

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Example Question #3 : Reactions And Titrations

Compound	Base strength, K_b
1	10¹²
2	10⁵
3	10¹
4	10^-8

Students in a chemistry class are given one of four unknown samples in a laboratory. A student is told that his compound is the strongest acid of the four compounds. Based on the information in the above table, which compound was the student given?

Possible Answers:

Compound 3

Compound 2

Compound 1

Compound 4

Correct answer:

Compound 4

Explanation:

The strongest acid of the group will also have the smallest K_b value. As the weakest base (smallest K_b), compound 4 will only partially dissociate in solution because it has a fairly strong conjugate acid. Were this question asking which base was the strongest, compound 1 would be the answer, due to its large K_b value.

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Example Question #4 : Reactions And Titrations

$HA \leftrightharpoons H^{+} + A^{-}$ $\displaystyle HA \leftrightharpoons H^{+} + A^{-}$

All of the bases proceed in a similar fashion.

$BOH \leftrightharpoons B^{+} + OH^{-}$ $\displaystyle BOH \leftrightharpoons B^{+} + OH^{-}$

In aqueous conditions the equilibrium constant for a Brønsted-Lowry base, $A^{-}$ $\displaystyle A^{-}$, can be expressed as which of the following?

Possible Answers:

$K_b=\frac{[HA][OH^-]}{[H_2O]}$ $\displaystyle K_b=\frac{[HA][OH^-]}{[H_2O]}$

$K_b=\frac{[OH^-]}{[A^-]}$ $\displaystyle K_b=\frac{[OH^-]}{[A^-]}$

$K_b=\frac{[HA][OH^-]}{[A^-]}$ $\displaystyle K_b=\frac{[HA][OH^-]}{[A^-]}$

$K_b=\frac{[A^-][OH^-]}{[HA]}$ $\displaystyle K_b=\frac{[A^-][OH^-]}{[HA]}$

$K_b=\frac{[HA]}{[A^-]}$ $\displaystyle K_b=\frac{[HA]}{[A^-]}$

Correct answer:

$K_b=\frac{[HA][OH^-]}{[A^-]}$ $\displaystyle K_b=\frac{[HA][OH^-]}{[A^-]}$

Explanation:

The base reaction will essentially be the reverse of the acid reaction. In the question, the aqueous conditions mean that the base, $A^{-}$ $\displaystyle A^{-}$, reacts with water to give the following reaction:

$A^{-} +H_{2}O \leftrightharpoons HA + OH^{-}$ $\displaystyle A^{-} +H_{2}O \leftrightharpoons HA + OH^{-}$

Following normal equilibrium convention, we omit water from the equation because it is a pure liquid.

$K_{eq}=\frac{[product]}{[reactant]}$ $\displaystyle K_{eq}=\frac{[product]}{[reactant]}$

$K_b=\frac{[HA][OH^-]}{[A^-]}$ $\displaystyle K_b=\frac{[HA][OH^-]}{[A^-]}$

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Example Question #1 : Reactions And Titrations

$H_{2}CO_{3} \left ( K_{a} = 4.5 * 10^{-7} \right )$ $\displaystyle H_{2}CO_{3} \left ( K_{a} = 4.5 * 10^{-7} \right )$

$HCHO_{2} \left ( K_{a} = 1.8 * 10^{-4} \right )$ $\displaystyle HCHO_{2} \left ( K_{a} = 1.8 * 10^{-4} \right )$

$HC_{2}H_{3}O_{2} \left ( K_{a} = 1.8 * 10^{-5} \right )$ $\displaystyle HC_{2}H_{3}O_{2} \left ( K_{a} = 1.8 * 10^{-5} \right )$

$HF \left ( K_{a} = 6.3 * 10^{-4} \right )$ $\displaystyle HF \left ( K_{a} = 6.3 * 10^{-4} \right )$

Given the above values of K_a, place the acids in order from strongest to weakest.

Possible Answers:

$HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$ $\displaystyle HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$

$H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2} > HF$ $\displaystyle H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2} > HF$

$HCHO_{2} > HF > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$ $\displaystyle HCHO_{2} > HF > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$

None of the above.

$HF > H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2}$ $\displaystyle HF > H_{2}CO_{3} > HC_{2}H_{3}O_{2} > HCHO_{2}$

Correct answer:

$HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$ $\displaystyle HF > HCHO_{2} > HC_{2}H_{3}O_{2} > H_{2}CO_{3}$

Explanation:

The acid dissociation constant, K_a, describes how strongly an acid tends to break apart into hydrogen ions (H⁺) and its conjugate base (A^-). The higher the dissociation constant, the stronger the acid. HF has the largest K_a of these acids, making it the strongest, and H₂CO₃ has the smallest K_a, making it the weakest.

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MCAT Physical : Acid-Base Chemistry

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Acid Base Equilibrium

Example Question #28 : Mcat Physical Sciences

Example Question #1 : Acid Base Chemistry

Example Question #1 : Acid Base Chemistry

Example Question #1 : Acid Base Chemistry

Example Question #2 : Reactions And Titrations

Example Question #1 : Acid Base Equilibrium

Example Question #3 : Reactions And Titrations

Example Question #4 : Reactions And Titrations

Example Question #1 : Reactions And Titrations

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