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High School Chemistry : High School Chemistry

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Example Questions

Example Question #1 : 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 100 times fewer 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.

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 #1 : P H

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

Possible Answers:

0.005M

0.03M

0.025M

0.0025M

0.003M

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 #2 : Calculating P H And P Oh

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

Possible Answers:

7.8

6.2

5.2

8.9

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

The pH of the solution cannot be determined

The pH of the solution is 11

The pH of the solution is 3

The pOH of the solution is 11

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

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:

0.25

1.78

1.03

4.0

6.23

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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Example Question #22 : Acids And Bases

You find a bottle in a lab that has a $10^{-5}M$ solution of acid. The acid has the following dissociative properties:

$HNO_3+H_2O\rightarrow H_3O^++NO_2^-$

What is the pH of this solution?

Possible Answers:

Correct answer:

Explanation:

HNO_3 is a strong acid, meaning it will completely dissociate in solution. As such, the concentration of the acid will be equal to the proton concentration. Thus, to find pH, you should just plug the molar concentration of the acid solution into the pH formula.

[HNO_3]=[H^+]

pH=-log[H^+]

$pH=-log(10^{-5})=5$

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Example Question #62 : Acid Base Chemistry

Calculate the pOH and hydroxide ion concentration in an organism whose blood pH was measured to be 8.00 at 25^oC .

Possible Answers:

pOH=1

$[OH^-]=8.0*10^{10}M$

pOH=1

$[OH^-]=6.0*10^{-10}M$

pOH=8

$[OH^-]=1.0*10^{-8}M$

pOH=6

$[OH^-]=1.0*10^{6}M$

pOH=6

$[OH^-]=1.0*10^{-6}M$

Correct answer:

pOH=6

$[OH^-]=1.0*10^{-6}M$

Explanation:

To find the pOH note that pH+pOH=14 . Therefore one can solve for pOH=14-pH . Plug in the value of the organism's blood pH and solve to get {pOH=6}

We find the concentration of hydroxide ions based on the formula for pOH:

pOH=-log[OH^-]

[OH^-]=antilog(-pOH) .

$[OH^-]=antilog(-6)=10^{-6}={1.0*10^{-6}M}$

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Example Question #61 : Acids And Bases

Calculate the pH of the following solution at 25^oC :

$1*10^{-5}M\ OH^-$

Possible Answers:

10.00

5.00

14.00

9.00

7.00

Correct answer:

9.00

Explanation:

Use the the dissociation constant for water $(K_{w})$ to calculate the concentration of hydrogen ions and then convert to the pH scale.

$K_{w}=[H^+][OH^-]=(1.0* 10^{-7})(1.0* 10^{-7})=1.0*10^{-14}$

Rearrange the dissociation constant to solve for the hydrogen ion concentration:

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

Plug in the given concentrations to find the actual hydrogen ion concentration:

$[H^+]=\frac{1.0* 10^{-14}}{1.0* 10^{-5}}=1.0*10^{-9}M$

Convert the hydrogen ion concentration to pH scale:

$pH=-log[H^+]=-log(1.0* 10^{-9}M)=9.00$

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Example Question #62 : Acids And Bases

Calculate the pOH of a .0750M HCl solution.

Possible Answers:

pOH = 8.3

pOH=7

pOH=12.9

pOH=1.1

pOH=.925

Correct answer:

pOH=12.9

Explanation:

Relevant equations:

pH=-log([H^+])

pOH=-log([OH^-])

pOH+pH=14

HCl is a strong acid, so [H^+]=[HCl]

Combine equations:

pOH-log([H^+])=14

Plug in values:

pOH-log([.0750])=14

pOH=12.9

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Example Question #1 : Concentration And Units

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

$\rho_{ethanol}=0.789\frac{g}{mL}$

Possible Answers:

94.7g

99.6g

92.8g

75g

102g

Correct answer:

94.7g

Explanation:

First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

$25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH$

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

$19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g$

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High School Chemistry : High School Chemistry

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #1 : Calculating P H And P Oh

Example Question #1 : P H

Example Question #2 : Calculating P H And P Oh

Example Question #11 : Calculating P H And P Oh

Example Question #61 : Acid Base Chemistry

Example Question #22 : Acids And Bases

Example Question #62 : Acid Base Chemistry

Example Question #61 : Acids And Bases

Example Question #62 : Acids And Bases

Example Question #1 : Concentration And Units

All High School Chemistry Resources

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