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High School Chemistry : Acids and Bases

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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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Example Question #13 : Calculating P H And P Oh

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 #61 : Acids And Bases

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=6

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

pOH=6

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

pOH=1

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

pOH=8

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

pOH=1

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

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 #11 : Calculating P H And P Oh

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

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

Possible Answers:

14.00

7.00

9.00

10.00

5.00

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 #461 : High School Chemistry

Calculate the pOH of a .0750M HCl solution.

Possible Answers:

pOH=12.9

pOH=1.1

pOH=7

pOH = 8.3

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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High School Chemistry : Acids and Bases

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #21 : Acids And Bases

Example Question #13 : Calculating P H And P Oh

Example Question #61 : Acids And Bases

Example Question #11 : Calculating P H And P Oh

Example Question #461 : High School Chemistry

All High School Chemistry Resources

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