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High School Chemistry : Acid-Base Chemistry

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Example Questions

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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Example Question #2 : Concentration And Units

What is the concentration if 40 g NaOH is dissolved in 1000 g of water

Possible Answers:

40 M

4 m

4 M

1 M

1 m

Correct answer: 1 M

Explanation:

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

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Example Question #3 : Concentration And Units

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

Possible Answers:

1.6 M

0.00004 M

0.0016 M

0.625 M

0.04 M

Correct answer:

0.04 M

Explanation:

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L =

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Example Question #4 : Concentration And Units

How many liters of 10M HCl are needed to make 4L of 0.4M solution?

Possible Answers:

0.16

8.0

1.6

0.8

0.4

Correct answer:

0.16

Explanation:

A simple calculation can be done to perform any solution dilution problem. We know our equation $molarity = \frac{moles}{volume}$ .

We can rewrite this as (molarity)(volume) = moles .

Using this formula, we take the old solution and set it equal to the new solution.

(10 M)(V) = (0.4M)(4L)=1.6mol

$V = \frac{1.6mol}{10M}=0.16L$

We need 0.16 liters of our 10 molar solution.

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Example Question #5 : Concentration And Units

Which solution has the highest molarity?

Possible Answers:

$\240g\ \text{of}\ MgSO_4\ \text{in}\ 3.0L\ H_2O$

$\138.0g\ \text{of}\ CH_3CH_2OH\ \text{in}\ 3.0L\ H_2O$

$\174g\ \text{of}\ NaCl\ \text{in}\ 2.0L\ H_2O$

$\85.0g\ \text{of}\ NaNO_3\ \text{in}\ 2.0L\ H_2O$

$\486.5g\ \text{of}\ Pb(NO_3)_2\ \text{in}\ 0.75L\ H_2O$

Correct answer:

$\486.5g\ \text{of}\ Pb(NO_3)_2\ \text{in}\ 0.75L\ H_2O$

Explanation:

This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.

Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.

$Pb(NO_3)_2:\ \frac{486.5g*\frac{1mol}{331g}}{0.75L}=1.96M$

$NaCl:\ \frac{174g*\frac{1mol}{58.5g}}{2.0L}=1.49M$

$MgSO_4:\ \frac{240g*\frac{1mol}{120.3g}}{3.0L}=0.67M$

$NaNO_3:\ \frac{85g*\frac{1mol}{85g}}{2.0L}=0.50M$

$CH_3CH_2OH:\ \frac{138g*\frac{1mol}{46g}}{3.0L}=1.0M$

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High School Chemistry : Acid-Base Chemistry

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

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

Example Question #2 : Concentration And Units

Example Question #3 : Concentration And Units

Example Question #4 : Concentration And Units

Example Question #5 : Concentration And Units

All High School Chemistry Resources

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