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High School Chemistry : Calculating pH and pOH

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Example Question #2 : P H

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.

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

Possible Answers:

The pH of the solution cannot be determined

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

The pH of the solution is 3

The pH of the solution is 11

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.

$\displaystyle pH + pOH = 14$

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

Use our value for the concentration to find the pOH.

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

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

$\displaystyle pH + 3 = 14$

pH = 11 $\displaystyle pH = 11$

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Example Question #1 : P H

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:

1.03 $\displaystyle 1.03$

1.78 $\displaystyle 1.78$

6.23 $\displaystyle 6.23$

0.25 $\displaystyle 0.25$

4.0 $\displaystyle 4.0$

Correct answer:

1.78 $\displaystyle 1.78$

Explanation:

The equation to calculate pH is:

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

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.

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

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

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

$\displaystyle 7.15=-log[H^+]$

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

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

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

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

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Example Question #11 : Calculating P H And P Oh

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

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

What is the pH of this solution?

Possible Answers:

$\displaystyle 8$

$\displaystyle 1$

$\displaystyle 5$

$\displaystyle 6$

$\displaystyle 2$

Correct answer:

$\displaystyle 5$

Explanation:

HNO_3 $\displaystyle HNO_3$ 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.

$\displaystyle [HNO_3]=[H^+]$

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

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

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Example Question #12 : Calculating P H And P Oh

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

Possible Answers:

pOH=1 $\displaystyle pOH=1$

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

pOH=8 $\displaystyle pOH=8$

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

pOH=6 $\displaystyle pOH=6$

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

pOH=1 $\displaystyle pOH=1$

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

pOH=6 $\displaystyle pOH=6$

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

Correct answer:

pOH=6 $\displaystyle pOH=6$

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

Explanation:

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

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

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

$\displaystyle [OH^-]=antilog(-pOH)$.

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

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Example Question #61 : Acids And Bases

Calculate the pH of the following solution at 25^oC $\displaystyle 25^oC$:

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

Possible Answers:

9.00 $\displaystyle 9.00$

7.00 $\displaystyle 7.00$

10.00 $\displaystyle 10.00$

14.00 $\displaystyle 14.00$

5.00 $\displaystyle 5.00$

Correct answer:

9.00 $\displaystyle 9.00$

Explanation:

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

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

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

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

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

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

Convert the hydrogen ion concentration to pH scale:

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

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Example Question #63 : Acids And Bases

Calculate the pOH $\displaystyle pOH$ of a .0750M $\displaystyle .0750M$ HCl $\displaystyle HCl$ solution.

Possible Answers:

pOH=7 $\displaystyle pOH=7$

$\displaystyle pOH=.925$

$\displaystyle pOH = 8.3$

pOH=1.1 $\displaystyle pOH=1.1$

$\displaystyle pOH=12.9$

Correct answer:

$\displaystyle pOH=12.9$

Explanation:

Relevant equations:

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

$\displaystyle pOH=-log([OH^-])$

$\displaystyle pOH+pH=14$

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

Combine equations:

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

Plug in values:

$\displaystyle pOH-log([.0750])=14$

$\displaystyle pOH=12.9$

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High School Chemistry : Calculating pH and pOH

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #2 : P H

Example Question #1 : P H

Example Question #11 : Calculating P H And P Oh

Example Question #12 : Calculating P H And P Oh

Example Question #61 : Acids And Bases

Example Question #63 : Acids And Bases

All High School Chemistry Resources

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