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MCAT Physical : Acid-Base Chemistry

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Example Question #11 : Acids And Bases

Which of the following is considered a weak base?

Possible Answers:

NaOH

NH_3

LiOH

KOH

Correct answer:

NH_3

Explanation:

Strong bases are those that completely dissociate in water. All group I hydroxides are considered strong bases, including sodium hydroxide, potassium hydroxide, and lithium hydroxide. Some group II hydroxides are also strong bases, including calcium hydroxide, barium hydroxide, and strontium hydroxide.

Ammonia is considered a weak base and is frequently used as a buffer in solution with ammonium, its conjugate acid.

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Example Question #12 : Acids And Bases

Which of the following is not a strong acid?

Possible Answers:

Acetic acid

Nitric acid

Hydrochloric acid

Perchloric acid

Correct answer:

Acetic acid

Explanation:

Strong acids are those that fully dissociate in an aqueous environment. Most protonated halogens are considered strong acids, with the exception of hydrogen fluoride; hydrochloric, hydrobromic, and hydroiodic acid are considered strong acids.

Perchloric acid is perhaps the strongest acid, and is formed from a hydrogen atom and a perchlorate ion: HClO_4 . Nitric acid is formed from a hydrogen atom and a nitrate ion, making it highly soluble and making it easy to dissolve in solution: HNO_3 .

Acetic acid, and most other organic acids, are considered weak acids since they lack stability when deprotonated. Stronger acids will form ions or resonace structures that are highly stable, favoring the release of the bound proton, while weaker acids will bind the proton more tightly to maintain a more stable structure. Acetic acid has the structure CH_3COOH .

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Example Question #13 : Acids And Bases

Which of the following is a weak base?

Possible Answers:

Ba(OH)₂

NH₃

K₂O

Li₂O

NaOH

Correct answer:

NH₃

Explanation:

Group 1 hydroxides, group 1 oxides, Ba(OH)₂, Sr(OH)₂, Ca(OH)₂, and metal amides are the only strong bases. Everything else is considered a weak base.

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Example Question #1 : Lewis, Brønsted Lowry, And Arrhenius Definitions

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.

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

All of the bases proceed in a similar fashion.

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

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.

In the reverse reaction of $HA \leftrightharpoons H^{+} + A^{-}$ , the proton is acting as a(n) __________, and is thus a __________.

Possible Answers:

electron acceptor . . . Lewis base

electron donor . . . Lewis acid

electron acceptor . . . Lewis acid

electron donor . . . Lewis base

electron donor . . . Brønsted-Lowry base

Correct answer:

electron acceptor . . . Lewis acid

Explanation:

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

In terms of the passage, the lone proton can be considered a proton donor and would, therefore, be a Brønsted-Lowry acid. This is not an answer choice.

The third acid-base definition is the Lewis definition, which states that acids are electron acceptors and bases are electron donors. The negative charge on the A^- signifies that it is a Lewis base with available electrons to donate. The proton is accepting these electrons from A^- , and is thus acting as a Lewis acid.

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Example Question #11 : Defining/Classifying Acids And Bases

The Haber-Bosch process, or simply the Haber process, is a common industrial reaction that generates ammonia from nitrogen and hydrogen gas. A worker in a company generates ammonia from the Haber process. He then dissociates the gaseous ammonia in water to produce an aqueous solution. Since ammonia is a base, it will accept a proton from water, generating OH^- and ammonium ion products. The two reactions involved are:

$N_2_(g_)+3H_2_(g_)\rightarrow2NH_3_(g_)\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)$

$NH_3_(g_)+H_2O_(l_)\rightarrow NH_4^+_(aq_)+OH^-_(aq_)\;\;\;\;\;(2)$

What happens when ammonium ions are dissociated in water?

Possible Answers:

Water gains a proton and generates hydroxide ions

Water is inert, so nothing happens

Water loses a proton and generates hydroxide ions

Water gains a proton and generates hydronium ions

Correct answer:

Water gains a proton and generates hydronium ions

Explanation:

Ammonia is a base and produces ammonium ions in water; therefore, ammonium is the conjugate acid of ammonia. This means that ammonium ions in water will donate protons to water, regenerating the conjugate base (ammonia). Protonated water contains a molecular formula of H_3O^+ , which are called hydronium ions.

$NH_4^++H_2O\rightleftharpoons NH_3+H_3O^+$

Remember that hydronium ions are always produced during acid dissociation. The presence of more hydronium ions indicates that the solution is acidic and has a low pH.

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Example Question #1 : P H

What is the resulting pH when 7.0g of HCl is dissolved in 3L of water?

Possible Answers:

2.8

1.6

1.2

0.7

2.1

Correct answer:

1.2

Explanation:

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 $\dpi{100} \small 6.7\times 10^{-2}$ .

Now we know that $\dpi{100} \small \left [ H^{+} \right ]=6.7\times 10^{-2}$ and pH=-log[H⁺]. Using our trick for -log, we can see that $\dpi{100} \small 1\times 10^{-1}> 6.7\times 10^{-2}> 1\times 10^{-2}$ . Since $\dpi{100} \small -log\left (1\times 10^{-1} \right )=1$ and $\dpi{100} \small -log\left (1\times 10^{-2} \right )=2$ , we know our answer is between 1 and 2.

$\dpi{100} \small 6.7\times 10^{-2}$ is closer to $\dpi{100} \small 1\times 10^{-1}$ , so we can pick the answer closer to 1. i.e. 1.2.

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Example Question #7 : Calculating P H And P Oh

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

All of the bases proceed in a similar fashion.

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

The pH of a solution of $\small HA$ is lowered from 4 to 3, and then from 3 to 2. Which of the following is the most accurate description of what happens during these transitions?

Possible Answers:

There are 20 times fewer protons in solution at pH 2 than at pH 4.

There are 100 times more protons in solution at pH 2 than at pH 4.

There are 20 times more protons in solution at pH 2 than at pH 4.

There are 20 times more hydroxide ions in solution at pH 2 than at pH 4.

There are 100 times fewer protons in solution at pH 2 than at pH 4.

Correct answer:

There are 100 times more protons in solution at pH 2 than at pH 4.

Explanation:

The pH scale is logarithmic. Every pH unit drop corresponds to a tenfold increase in protons.

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

$2=-\log(0.01)$

$4=-\log(0.0001)$

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Example Question #8 : Calculating P H And P Oh

A sample of gastric juice has a pH of 2.5. What is the hydrogen ion concentration in this secretion?

Possible Answers:

0.025M

0.003M

0.0025M

0.03M

0.005M

Correct answer:

0.003M

Explanation:

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

The concentration of hydrogen ions must lie somewhere between $\small 10^{-3}M$ and $\small 10^{-2}M$ ; alternatively stated, it is between $\small 1*10^{-3}M$ and $\small 10*10^{-3}M$ . The pH of a solution with hydrogen ion concentration of $\small 10^{-3}M$ will be 3, and the pH of a solution with hydrogen ion concentration $\small 10^{-2}M$ will be 2; thus, our concentration must lie between these two values, since our pH is 2.5

2.5=3.0-0.5

To find the exact concentration, you must be familiar with the logarithmic scale. A difference of 0.5 is equivalent to a log of 3.

$\log(3)=0.48$

Our answer must therefore be $\small 3*10^{-3}M$ , or $\small 0.003M$ .

We can calculate the pH in reverse to check our answer.

$-\log(3*10^{-3})$

$-(\log(3)+\log(10^{-3}))$

-(0.48+(-3))

-(-2.52)

2.52

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Example Question #11 : Acids And Bases

2.0g of a monoprotic strong acid are dissolved in 1L of water. The resulting pH is 2.0. What is the molecular weight of the acid?

Possible Answers:

$\dpi{100} \small \frac{225g}{mol}$

$\dpi{100} \small \frac{100g}{mol}$

$\dpi{100} \small \frac{200g}{mol}$

$\dpi{100} \small \frac{150g}{mol}$

the answer cannot be determined

Correct answer:

$\dpi{100} \small \frac{200g}{mol}$

Explanation:

Using the pH of 2.0, we can find that $\dpi{100} \small \left [ H^{+} \right ]=10^{-2}$ , because pH=-log[H^+] .

Since the acid is strong (fully dissociates) and monoprotic (one H⁺per molecule), $\small [acid]=[H^+]=10^{-2}$ .

Our solution has 1L of water, meaning that we have $\dpi{100} \small 10^{-2}mole$ of acid. We know that only 2.0g of acid were used to achieve this concentration, meaning that there is a ratio of $\dpi{100} \small \frac{2.0g}{10^{-2}mole}$ . Simplifying this ratio gives $\dpi{100} \small \frac{200g}{mol}$ .

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Example Question #1 : Calculating P H And P Oh

What is the pOH of a $\small 6*10^{-7}M$ aqueous solution of $\small HCl$ ?

Possible Answers:

5.2

8.9

6.2

7.8

Correct answer:

7.8

Explanation:

The first step for this problem is to find the pH. We can then derive the pOH from the pH value.

The pH is given by the equation $pH=-\log[H^+]$ . Since hydrochloric acid is monoprotic, the concentration of the solution is equal to the concentration of protons.

$[HCl]=[H^+]=6*10^{-7}M$

Using this value and the pH equation, we can calculate the pH.

$pH=-\log(6*10^{-7})=6.2$

Now we can find the pOH. The sum of the pH and the pOH is always 14.

pH+pOH=14

pOH=14-pH

pOH=14-6.2=7.8

The pOH of the solution is 7.8.

Alternatively, a shortcut can be used to estimate the pH. If $\small [H^+]$ is in the form $\small a*10^{-b}$ , then pH is roughly (b-1).(10-a) .

(7-1).(10-6)=6.4

For this question, this shortcut gets us a pH of 6.4, which produces a pOH of 7.6; very close to the real answer!

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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 #11 : Acids And Bases

Example Question #12 : Acids And Bases

Example Question #13 : Acids And Bases

Example Question #1 : Lewis, Brønsted Lowry, And Arrhenius Definitions

Example Question #11 : Defining/Classifying Acids And Bases

Example Question #1 : P H

Example Question #7 : Calculating P H And P Oh

Example Question #8 : Calculating P H And P Oh

Example Question #11 : Acids And Bases

Example Question #1 : Calculating P H And P Oh

All MCAT Physical Resources

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