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High School Chemistry : Acid-Base Chemistry

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Example Questions

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Example Question #6 : Titrations

Which of the following aqueous solutions is the most concentrated?

$\small \rho_{water}=1 \frac{g}{mL}$

Possible Answers:

$\small \small 1m HCl$

$36.5g\ \text{HCl in}\ 1kg\ \text{water}$

$\small \small 1M HCl$

All of these solutions have the same concentration.

Correct answer:

$\small \small 1M HCl$

Explanation:

In order to answer this question, it helps to know that 1 kilogram of water is equal to 1 liter of water, due to its density. Two of the above options refer to a 1m solution of hydrochloric acid. The other is a 1M solution.

$1M\ HCl=\frac{1mol\ HCl}{1L}$

$1m\ HCl=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

$36.5g\ \text{HCl in}\ 1kg\ \text{water}=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

All three of the options have the same amount of hydrochloric acid (one mole). For molarity, the hydrochloric acid is diluted with water until one liter of solution is created. For molality, one mole of HCl is added to one kilogram of water. Since one kilogram of water is one liter, this becomes the same concentration.

One a very small level, the 1M HCl solution will be slightly more concentrated. Creating a molal solution does not take into account the volume of the solute. If, for example, 100 cubic centimeters of HCl were added to one kilogram of water, the resulting volume would be more than one liter, making the concentration slightly less than 1M. This discrepancy is usually not accounted for in basic chemistry, but you should be familiar with the concept.

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Example Question #1 : Concentration And Units

What is the molality of a solution created by mixing 4.3g $NaCl$ into 43g of water?

Possible Answers:

0.03m

0.10m

0.71m

1.71m

Cannot be determined

Correct answer:

1.71m

Explanation:

Molality can be defined:

$Molality=\hspace{1 mm}\frac{mol\hspace{1 mm}solute}{kg\hspace{1 mm}solvent}$

It is slightly different from Molarity and has different uses.

$Molality=\hspace{1 mm}\frac{4.3\hspace{1 mm}g\hspace{1 mm}NaCl}{43\hspace{1 mm}g\hspace{1 mm}H_20}\times\frac{1000\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}kg\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaCl}{58.44\hspace{1 mm}g\hspace{1 mm}NaCl}=1.71\hspace{1 mm}m$

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Example Question #1 : Identifying Unknown Concentration

How much solid NaOH must be dissolved to make 740mL of a 0.32M solution?

Possible Answers:

9.47g

9.47 * 10²g

The sodium hydroxide will boil off with the water

4.26g

12.8g

Correct answer:

9.47g

Explanation:

This problem can be solved by stoichiometry. Remember that 0.32M gives us the moles of NaOH per liter, and solve for the number of moles per 0.740L.

$740\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.32\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=9.47\hspace{1 mm}g\hspace{1 mm}NaOH$

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Example Question #9 : Titrations

Find the mass of KOH in 10L of water if it is a 2m solution.

Possible Answers:

112g

560g

224g

2240g

1120g

Correct answer:

1120g

Explanation:

Molality is grams of solute per kilogram of solvent.

$m=\frac{mol}{kg}\rightarrow mol=(m)(kg)$

Water has a density of one gram per mililiter, so one liter of water equal to one kilogram. If we have a 2m solution, that means we have two moles of KOH per kilogram of water.

$10L*\frac{1kg}{1L}=10kg$

mol=(m)(kg)=(2m)(10kg)=20mol

KOH has a molecular weight of $\small 56\frac{g}{mol }$ .

$20mol*\frac{56g}{1mol}=1120g\ KOH$

This gives us 1120 g of KOH .

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Example Question #10 : Titrations

10mL of a solution of NaOH of unknown concentration mixed with 34mL of $10^{-4}M$ HCl produced a solution with a pH of 7. What is the concentration of the NaOH solution?

Possible Answers:

None of these

$3.4\cdot 10^{-4}M$

3.4M

0.01M

0.001M

Correct answer:

$3.4\cdot 10^{-4}M$

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

N_A is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=3.4\cdot 10^{-4}M$

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Example Question #11 : Identifying Unknown Concentration

30mL of a solution of NaOH of unknown concentration mixed with 19mL of $10^{-3}M$ HCl produced a solution with a pH of 7. What is the concentration of the NaOH solution?

Possible Answers:

$6.3\cdot 10^{-3}$

0.002

None of these

0.0001

$6.3\cdot 10^{-4}$

Correct answer:

$6.3\cdot 10^{-4}$

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=6.3\cdot10^{-4}M$

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Example Question #72 : Acid Base Chemistry

29mL of a solution of $10^{-4}M$ NaOH is mixed with 65mL of an HCl solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the HCl solution?

Possible Answers:

$1.5\cdot10^{-5}M$

$10^{-4}M$

None of these

$4.5\cdot10^{-5}M$

$3.0\cdot10^{-5}M$

Correct answer:

$4.5\cdot10^{-5}M$

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

$M_A=4.5\cdot10^{-4}M$

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Example Question #11 : Titrations

39mL of a solution of $10^{-4}M$ NaOH is mixed with 24mL of an HCl solution of unknown concentration produced a solution with a pH of 7. What is the concentration of the HCl solution?

Possible Answers:

$1.5\cdot10^{-4}$

$2.5\cdot10^{-4}$

None of these

$1.63\cdot10^{-4}M$

$3.75\cdot10^{-4}$

Correct answer:

$1.63\cdot10^{-4}M$

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

$M_A=1.63\cdot10^{-4}M$

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Example Question #11 : Titrations

It takes 25mL of a .001M NaOH

to neutralize 5mL of a solution of HCl

of unknown concentration. What is the concentration of the unknown solution?

Possible Answers:

.005M

.010M

.015M

.05M

None of these

Correct answer:

.005M

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the molarity of the acid:

$\frac{N_B M_B V_B }{N_A V_A}= M_A$

Plug in known values and solve.

M_A=0.005M

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Example Question #71 : Acid Base Chemistry

It takes 35mL of .01M HCl

to neutralize 5mL of a solution of NaOH

of unknown concentration. What is the concentration of the unknown solution?

Possible Answers:

.07M

.007M

.014M

None of these

.035M

Correct answer:

.07M

Explanation:

For this question use the following formula:

is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the acid:

Plug in known values and solve.

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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 #6 : Titrations

Example Question #1 : Concentration And Units

Example Question #1 : Identifying Unknown Concentration

Example Question #9 : Titrations

Example Question #10 : Titrations

Example Question #11 : Identifying Unknown Concentration

Example Question #72 : Acid Base Chemistry

Example Question #11 : Titrations

Example Question #11 : Titrations

Example Question #71 : Acid Base Chemistry

All High School Chemistry Resources

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