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High School Chemistry : High School Chemistry

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Example Questions

Example Question #481 : High School 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:

$4.5\cdot10^{-5}M$

$10^{-4}M$

$3.0\cdot10^{-5}M$

None of these

$1.5\cdot10^{-5}M$

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

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 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:

$2.5\cdot10^{-4}$

$3.75\cdot10^{-4}$

$1.63\cdot10^{-4}M$

$1.5\cdot10^{-4}$

None of these

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 #482 : High School Chemistry

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

.015M

None of these

.05M

.010M

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 #12 : Titrations

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:

.014M

.07M

.035M

.007M

None of these

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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Example Question #1 : Identifying Unknown Volume

What volume of 0.5M sulfuric acid is needed to neutralize 50mL of 0.5M potassium hydroxide?

Possible Answers:

75mL

100mL

12.5mL

50mL

25mL

Correct answer:

25mL

Explanation:

Sulfuic acid is a diprotic acid and will give two hydrogen ions for each molecule of sulfuric acid that is completely dissociated. Potassium hydroxide has only one hyroxide ion per molecule.

$H_2SO_4\rightleftharpoons 2H^++SO_4^{2-}$

$KOH\rightleftharpoons K^++OH^-$

Both of these compounds are strong, meaning they will dissociate completely in solution (potassium hydroxide is a strong base and sulfuric acid is a strong acid).

$H_2SO_4+2KOH\rightleftharpoons 2H_2O+K_2SO_4$

First, we need to find the moles of potassium hydroxide from the given volume and molarity.

$M=\frac{mol}{L}\rightarrow mol=(M)(L)$

$mol=(0.5M)(50mL)=(0.5M)(0.05L)=0.025mol\ KOH$

Use stoichiometry to find the volume of sulfuric acid needed to react 0.1mol of potassium hydroxide.

$0.025mol\ KOH*\frac{1mol\ H_2SO_4}{2mol\ KOH}*\frac{1L}{0.5mol\ H_2SO_4}=0.025L\ H_2SO_4$

Our final volume of sulfuric acid is 0.025L, or 25mL.

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Example Question #22 : Solutions

A chemist has a bottle containing a 2M aqueous solution of hydrochloric acid. He needs to create a 50mL solution of hydrochloric acid that has a concentration of 0.5M. What is the volume of 2M hydrochloric acid that he should dilute in order to achieve the desired concentration?

Possible Answers:

$\small 1.25mL$

$\small 37.5mL$

$\small 5mL$

$\small 12.5mL$

Correct answer:

$\small 12.5mL$

Explanation:

In order to dilute the concentrated acid, we need to find the amount of concentrated acid that will be diluted to 50mL of total solution. We can find the volume of concentrated acid necessary by setting the final volume and concentration equal to the initial concentration and unknown volume.

$\small M_{1}v_{1} = M_{2}v_{2}$

The initial concentration is 2M, the final concentration is 0.5M, and the final volume is 50mL

$\small (2M)v_{1} = (0.5M)(50 mL)$

$\small v_{1} = \frac{(0.5M)(50mL)}{2M}=12.5 mL$

This means that 12.5mL of concentrated acid needs to be diluted to 50mL of solution. This will result in a solution with a concentration of 0.5M.

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Example Question #3 : Identifying Unknown Volume

Starting with 30mL of a .005M HCl solution, how many milliliters of .010M NaOH solution are needed to titrate it?

Possible Answers:

15mL

30mL

None of these

45mL

10mL

Correct answer:

15mL

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 volume of the base:

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

Plug in known values and solve.

V_B=15mL

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Example Question #4 : Identifying Unknown Volume

Starting with 50mL of a .015M HNO_3 solution, how many milliliters of .010M NaOH solution are needed to titrate it?

Possible Answers:

75mL

50mL

10mL

25mL

None of these

Correct answer:

75mL

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 volume of the base:

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

Plug in known values and solve.

V_B=75mL

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Example Question #14 : Titrations

Starting with 25mL of a .001M HNO_3 solution, how many milliliters of .010M Mg(OH)_2 solution are needed to titrate it?

Possible Answers:

1.25mL

None of these

50mL

2.5mL

25mL

Correct answer:

1.25mL

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 volume of the base:

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

Plug in known values and solve.

V_B=1.25mL

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Example Question #1 : Identifying Unknown Volume

Starting with 75mL of a .030M HNO_3 solution, how many milliliters of .022M NaOH solution are needed to titrate it?

Possible Answers:

102mL

25mL

50mL

75mL

None of these

Correct answer:

102mL

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 volume of the base:

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

Plug in known values and solve.

V_B=102mL

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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 #481 : High School Chemistry

Example Question #11 : Titrations

Example Question #482 : High School Chemistry

Example Question #12 : Titrations

Example Question #1 : Identifying Unknown Volume

Example Question #22 : Solutions

Example Question #3 : Identifying Unknown Volume

Example Question #4 : Identifying Unknown Volume

Example Question #14 : Titrations

Example Question #1 : Identifying Unknown Volume

All High School Chemistry Resources

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