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MCAT Physical : 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

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.

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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Example Question #3 : P H

47.0g of nitrous acid, HNO₂, is added to 4L of water. What is the resulting pH? $\dpi{100} \small \left (K_{a}=4.1\times 10^{-4} \right )$

Possible Answers:

3.2

2.5

3.5

3.0

2.0

Correct answer:

2.0

Explanation:

HNO₂ is a weak acid; it will not fully dissociate, so we need to use the HA → H⁺+ A^– reaction, with $\dpi{100} \small K_{a}=\frac{\left [ products \right ]}{\left [ reactants \right ]}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=4.1\times 10^{-4}$ .

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: $\dpi{100} \small K_{a}=\left \frac{\left [ H^{+} \right ]\left [ A^{-} \right ]}{\left [ HA \right ]}=\frac{\left ( x \right )\left ( x \right )}{0.25-x}$ .

Since x is very small, we can ignore it in the denominator: $\dpi{100} \small K_{a}=\left \frac{\left ( x \right )\left ( x \right )}{\left (0.25 \right )}=4.1\times 10^{-4}$

(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 $x=1\times10^{-2}$ . Looking at our table, we know that $\dpi{100} \small x=\left [ H^{+} \right ]$

Now we can solve for pH: $pH=-log[H^+]=-log(1\times10^{-2})=2.0$

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Example Question #11 : Calculating P H And P Oh

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

All of the bases proceed in a similar fashion.

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

A scientist is studying an aqueous sample of A^- , and finds that the hydroxide concentration is $\small 10^{-3} M$ . Which of the following is true?

Possible Answers:

The pH of the solution is 11

The pH of the solution cannot be determined

The concentration of protons is $\small 10^{-9}M$

The pOH of the solution is 11

The pH of the solution is 3

Correct answer:

The pH of the solution is 11

Explanation:

Given the hydroxide ion concentration, we will need to work using pOH to find the pH. We know that the sum of pH and pOH is equal to 14.

pH + pOH = 14

$pOH=-\log[OH^-]$

Use our value for the concentration to find the pOH.

$pOH=-\log(10^{-3})=3$

Now that we have the pOH, we can use it to solve for the pH.

pH + 3 = 14

pH = 11

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Example Question #61 : Acid Base Chemistry

HCN dissociates based on the following reaction.

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

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

0.4mol of HCN is added to of water. What is the pH of the resulting solution?

Possible Answers:

2.54

9.91

4.95

0.7

3.77

Correct answer:

4.95

Explanation:

Since HCN is a weak acid, we must use the equilibrium equation.

$K_{a}= \frac{[CN^{-}][H^{+}]}{[HCN]}$

Because the HCN dissociates in solution, we expect the concentrations of protons and cyanide ions to increase, while the concentration of HCN will decrease. After determining the molarity of the solution, we can set up the equation below, using X as the amount of moles that dissociate.

$\small K_{a}= \frac{[X][X]}{[[\frac{0.4mol}{2L}]-X]} = 6.2*10^{-10}$

Because X is small, we can neglect its impact in the denominator.

$\small K_{a}= \frac{[X][X]}{[\frac{0.4mol}{2L}]} = \frac{[X]^2}{[0.2]}=6.2*10^{-10}$

$\small X = 1.1*10^{-5}$

Since X is the concentration of protons in the solution, we can calculate the pH by using the equation $pH = -log[H^{+}]$ .

$pH=-log[1.1*10^{-5}]=4.95$

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Example Question #5 : P H

There are two containers, each containing different proton concentrations. In container A, $[H^{+} ] = 5 *10^{-6}M$ . In container B, $[H^{+}] = 1*10^{-3}$ .

What is the difference in pH between these two containers?

Possible Answers:

1.4

3.2

2.3

4.6

0.9

Correct answer:

2.3

Explanation:

The MCAT will typically allow you to approximate the difference in pH between two solutions. Look at the problem this way: you should already know a proton concentration of 1 * 10^-3M means a pH of 3. You do not need to know the exact pH of 5 * 10^-6M, but you should recognize that it is in between 1 * 10^-6M and 1 * 10^-5M. This means it will have a pH between 5 and 6.

The difference in pH will therefore be between 5 minus 3 and 6 minus 3.

$pH\ difference=pH_A-pH_B$

5<pH_A<6

$5-3<pH\ difference<6-3$

$2<pH\ difference<3$

As a result, look for the answer that is between 2 and 3. The only answer that makes sense is 2.3.

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Example Question #21 : Acids And Bases

An arterial blood sample from a patient has a pH of 7.4. One day later, the same patient has an arterial blood pH of 7.15. How many times more acidic is the patient's blood on the second day?

Possible Answers:

1.78

6.23

4.0

1.03

0.25

Correct answer:

1.78

Explanation:

The equation to calculate pH is:

$\small pH = - log$ $\small \left [ H^{+}\right ]$

The normal pH of arterial blood is around 7.4. This reflects a concentration of hydrogen ions that can be found using the pH equation.

$\small 7.4 = - log\left [ H^{+}\right ]$

$\small 3.98 *10^{-8}M = \left [ H^{+}\right ]$

Using similar calculations for the second blood sample, we can find the hydrogen ion concentration again.

7.15=-log[H^+]

$7.08*10^{-8}M=[H^+]$

Now that we have both concentrations, can find the ratio of the acidity of the two samples.

$\frac{[H^+_2]}{[H^+_1]}=\frac{7.08*10^{-8}}{3.98*10^{-8}}=1.78$

You may know from biological sciences that this is approaching a lethal level of acidosis.

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MCAT Physical : pH

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

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

Example Question #3 : P H

Example Question #11 : Calculating P H And P Oh

Example Question #61 : Acid Base Chemistry

Example Question #5 : P H

Example Question #21 : Acids And Bases

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