Solving Weak Acid/Base Dissociation Problems

Dissociation problems are about calculations of hydronium ion concentration, [H₃O⁺], and/or pH of aqueous solutions containing any of the following:

a weak acid or a weak base

a strong acid or a strong base

a salt of a weak acid or a salt of a weak base

a weak acid and its salt (weak acid/conjugate base composition) or a weak base and its salt (weak base/conjugate acid composition)

Weak acid and weak base dissociate (or ionize) partially in aqueous solution. Their dissociation products combine to form again the initial weak acid and base.These two reactions occur simultaneously and form an equilibrium. This is in contrast to the strong acids and bases that undergo complete dissociation or ionization in aqueous solution.

Basically, these reactions are about giving up and gaining protons. Those that donate protons are called Bronsted-Lowry acids; and those that accept protons are called Bronsted-Lowry bases.

Their dissociation equilibria given below show that a weak acid loses a proton and forms conjugate base while a weak base gains a proton and forms conjugate acid.

The extent of dissociation (or ionization) is determined by their so called dissociation constants ( K_a for acid and K_b for base ), measured at 25° C.

The dissociation constant is a measure of the tendency of a weak acid to give up a proton and tendency of a weak base to gain a proton. A very small dissociation constant value indicates a very low degree of dissociation.

Values in [ ] denote concentration in moles/liter.

Since the [H₂O] in each case is considered constant, it is incorporated into K_a and K_b.

Take note here that water acts as a weak base in weak acid dissociation and acts as a weak acid in weak base dissociation. A solvent that can act either as an acid or as a base depending upon the solute is called amphiprotic ( or amphoteric ) solvent.

The following discussion is about the calculations of hydronium ion concentration of the aqueous solutions containing (a) a weak acid or a weak base, (b) a strong acid or a strong base, (c) a salt of a weak acid or a salt of a weak base, and (d) a weak acid and its salt or a weak base and its salt.

A. Aqueous Solution of a Weak Acid or a Weak Base

The [H₃O⁺] and/or pH of this solution can be calculated using equations 1 and 2.

Since the amount (or concentration) of dissociated acid or base is equal to the amount (or concentration) of dissociation products:

[dissociated acid] = [H₃O⁺] = [A^-] = x, and

[ionized base] = [BH⁺] = [OH^-] = x, then

we can set equations 1 and 2 to one unknown:

Equations 3 and 4 have the general form of ax² + bx + c = 0 which can be solved by using the quadratic equation:

If the dissociation constant value is very low such that x is negligible, the following approximation can be made:



If the dissociation constant value or % error is high, equations 3 and 4 should then be used for calculation.

B. Aqueous Solution of a Strong Acid or a Strong Base

At 25° C, water has the following dissociation equilibrium:

Since [H₂O] is considered constant, it is incorporated into K_w.

K_w is called the ion-product constant of water at 25° C. This means that pure water has this much amount dissociated at this temperature, that is, [H₃O⁺] = [OH^-] = 1 x 10^-7 M. The [H₃O⁺] or [OH^-] of an aqueous solution can be calculated when one of them is known.

Addition of a strong acid or base in water will shift the equilibrium to the left, as predicted by Le Chatelier principle, lowering the [H₃O⁺] and [OH^-] by depressing the dissociation of water.

Since a strong acid or base dissociates completely in water, then:

[acid] = [H₃O⁺]

[base] = [OH^-]

So,

Setting equation 7 to one unknown:

[dissociated water] = [H₃O⁺] = [OH^-] = x

1 x 10^-14 = ( [acid or base] + x )( x )

which can be solved by using quadratic equation.

But x is very small, we can assume that:

The sample problem below shows the validity of our assumption that x is negligible.

C. Aqueous Solution of a Salt of a Weak Acid or a Salt of a Weak Base

Salt of a weak acid dissociates completely in water as Na⁺ and A^-.

A^-, as a conjugate base of HA, reacts with water to form the following equilibrium:

Multiplying K_a (equation 1) by K_b, we get:



Equation 8 indicates the relationship that the strength of a weak acid or base tends to increase as the strength of its conjugate pair decreases, or vice versa.

This equation enables us to compute the dissociation constant value of the conjugate base of a weak acid and the conjugate acid of a weak base, as well as the [H₃O⁺] of this solution.

D. Aqueous Solution of a Weak Acid And Its Salt or a Weak Base And Its Salt

An aqueous solution containing a weak acid and its salt (or a weak base and its salt) is called a buffered solution or simply buffer solution. This is so because a buffer solution's hydronium ion concentration changes very little upon its dilution or upon addition of a strong acid/base.
The changes in pH value are so small such that the pH of the solution remains practically constant.

Calculation of the buffer solution's pH involves the use of equation 1 or 2:

By taking the negative logarithm of each term of the rearranged equation 1, we get equation 9 which is known as the Henderson-Hasselbach equation.



This equation is commonly used for pH calculation in biochemistry problems.

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Problems

The best way to improve your problem solving skill in chemistry is to practice solving as many chemistry problems as possible. Try solving the following problems.

Problems are solved using approximation and the quadratic equation. Values in parenthesis are obtained using the quadratic equation.

For more problems on this topic and other chemistry topics, go to www.tutorpartner.blogspot.com.

1.	Calculate the dissociation constant of the conjugate base of HC2H2ClO2. Ka = 1.36 x 10-3.
	Answer: 7.35 x 10-12
2.	A one liter aqueous solution contains 9.54 g KOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( KOH = 56.1056 g )
	Answer: pH = 13.23
3.	An aqueous solution has a strength of 0.05 M HOCN and 0.07 M NaOCN. Calculate the pH of this solution. ( Ka = 3.3 x 10-4)
	Answer: pH = 3.63 ( 3.63 )
4.	Calculate the pH of a 0.3 M NaC6H5O solution. (Ka for phenol acid is 1.05 x 10-10; Kw = 1 x 10-14)
	Answer: pH = 11.73 ( 11.72 )
5.	110 mL of 0.01 M NaOH solution is added to 110 mL of 0.11 M HNO2 solution. Find the pH of the resulting solution. Given: Ka for nitrous acid is 4.5 x 10-4.
	Answer: pH = 2.35 ( 2.56)
6.	A one liter aqueous solution contains 9.2 g NaOH. What is the pH of this solution? The dissociation constant for water, Kw, is 1 x 10-14. ( NaOH = 39.99707 g )
	Answer: pH = 13.36
7.	What is the pH of an aqueous solution having a concentration of 0.07 M HOCN and 0.02 M NaOCN? ( Ka = 3.3 x 10-4)
	Answer: pH = 2.94 ( 2.97 )
8.	If a solution of HOCN has a concentration of 0.9 M, what is its pH? Ka for cyanic acid is 3.3 x 10-4.
	Answer: pH = 1.76 ( 1.77 )
9.	Find the pH of a 0.86 M NaC6H5COO solution. (Ka for benzoic acid is 6.3 x 10-5; Kw = 1 x 10-14)
	Answer: pH = 9.07 ( 9.07 )
10.	An aqueous solution of 0.12 M CH3COOH and 0.12 M NaCH3COO has a volume of 340 mL. If 1.6 g of NaOH is added to this solution, find the change in pH. Assume no change in the volume of solution. ( Ka = 1.8 x 10-5; NaOH = 39.99707 g )
	Answer: pH = 6.75 ( 6.75)
11.	What is the pH of a solution of 0.28 M NaCH3COO? Calculate the percent hydrolysis. (Ka = 1.8 x 10-5; Kw = 1 x 10-14)
	Answer: pH = 9.1; 4.45 x 10-3 % ( 9.1; 4.45 x 10-3 % )
12.	What is the pH of an aqueous solution of 0.09 M C6H5NH3Cl? ( Given: Kb = 3.94 x 10-10; Kw = 1 x 10-14)
	Answer: pH = 2.82 ( 2.82 )
13.	If 3 L of 0.01 M HOC2H4NH2 solution has a pH of 10.37, find the weight of HOC2H4NH3Cl dissolved in the solution. ( Kb for ethanolamine is 3.18 x 10-5; mol. wt. = 97.5443 )
	Answer: 0.4 g
14.	In a buffer solution of 0.07 M HC2H2ClO2 and 0.07 M NaC2H2ClO2, HCl is added to it such that its calculated concentration in the solution is 0.02 M. Assuming a constant volume of solution, find the change in pH of the solution. ( Ka = 1.36 x 10-3)
	Answer: pH = 2.61 ( 2.64 )
15.	Find the [H+] and [OH-] of a 0.24 M of HF solution. Ka = 6.7 x 10-4; Kw = 1 x 10-14.
	Answer: [H+] = 1.27 x 10-2 M; [OH-] = 7.89 x 10-13 M ( 1.24 x 10-2; 8.1 x 10-13 )
16.	What is the dissociation constant of the conjugate acid of HOC2H4NH2. Kb = 3.18 x 10-5.
	Answer: 3.14 x 10-10
17.	What is the pH of a 0.38 M CH3COOH solution. ( Ka = 1.8 x 10-5 )
	Answer: pH = 2.58; ( 2.58)
18.	What is the pH of an aqueous solution of NH2C2H4NH2 having a concentration of 0.01 M? The dissociation constant of ethylenediamine is 8.5 x 10-5.
	Answer: pH = 10.96 ( 10.94)

Drawing Lewis Electron Dot Structure or Formula

A Lewis electron dot structure shows how the atoms of a molecule or an ion share their outermost electrons or valence electrons to form covalent bonds with each other.

It indicates the number of shared and unshared valence electrons.

This sharing of electrons results in their outermost shells being filled up thus attaining electronic configuration similar to those of the noble gases'.

Observe the electronic configurations of the noble gases below.

He	1s2
Ne	1s22s22p6
Ar	1s22s22p63s23p6
Kr	1s22s22p63s23p64s23d104p6
Xe	1s22s22p63s23p64s23d104p65s24d105p6
Rn	1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p6

You can see that their outermost shells are filled up which account for their being highly stable.

Now, compare them to the electronic configurations of several atoms that commonly form covalent bonds with other atoms.

H	1s1
C	1s22s22p2
O	1s22s22p4
N	1s22s22p3
S	1s22s22p63s23p4
B	1s22s22p1
P	1s22s22p63s23p3
Si	1s22s22p63s23p2
F	1s22s22p5

With the exception of hydrogen, they share their valence electrons with other atoms to fill up their outermost shells which contain a maximum of eight electrons. Hence, a Lewis electron dot formula must follow the octet (eight) rule.

Take a look at the following examples.

For our purposes, we're going to use color codes in our illustrations to distinguish the individual atoms with their respective valence electrons.

Lewis Electron Dot Structure of Ethane, C₂H₆

The electronic configuration of carbon atom shows that it needs four more electrons to fill up its outer shell; the hydrogen atom, on the other hand, needs just one more electron for its lone electron shell.

Each carbon atom satisfies the octet rule by sharing 3 valence electrons with 3 hydrogen atoms and one with another carbon atom. The hydrogen atom fills up its shell by sharing one electron with carbon atom.

Take note that each pair of shared electrons constitue a single bond.

Lewis Electron Dot Structure of Ethylene (Ethene), C₂H₄

This time, the carbon atoms share 2 electrons with each other and 2 electrons with 2 hydrogen atoms.

Sharing of two pairs of electrons constitute double bond.

Lewis Electron Dot Structure of Acetylene (Ethyne), C₂H₂

Here the carbon atoms share 3 valence electrons with each other and 1 electron with 1 hydrogen atom.

Sharing of three pairs of electrons constitute triple bond.

So far, we've been looking at the Lewis electron dot structures of molecules. We're going to look at the electron dot structures of some ions.

Lewis Electron Dot Structure of Hydroxide Ion

From the electronic configurations given above, we can see that oxygen atom has 6 valence electrons; it needs 2 more electrons to fill up its outer electron shell.

In order to follow the octet rule, oxygen forms the hydroxide ion with the hydrogen atom by sharing one electron with one hydrogen atom and acquiring one more electron .

The gain of one electron results to the ion having a negative net charge of one.

The structure has one electron more than the total valence electrons of the ion (6 from O + 1 from H = 7 valence electrons).

Lewis Electron Dot Structure of Ammonium Ion

The nitrogen atom has 5 valence electrons as indicated by its electronic configuration; it needs three more electrons to fill up its last shell.

But the nitrogen atom in the ammonium ion satisfies the octet rule by sharing three of its valence electrons with each of the three hydrogen atoms.

The structure has one electron less than the total valence electrons of the ion (5 from N + 4 from 4 H = 9 valence electrons).

Determining The Net Charge of an Electron Dot Structure

From these examples, we can say that the:

positive net charge of an ion is equal to the number of electrons deficient in its total valence electrons;

negave net charge of an ion is equal to the number of electrons in excess of its total valence electrons.

The net charge of an ion is also determined by the algebraic sum of the formal charge of all the atoms of the ion.

The formal charge of an atom is the encircled + or - sign near the atom. It is calculated as follows:

formal charge of an atom = valence electrons of the atom - (number of unshared electrons of the atom + number of pairs of shared electrons by the atom)

Formal charge of hydroxide ion's atoms:

O = 6 - (6 + 1) = -1

H = 1 - (0 + 1) =   0

net charge of hydroxide ion = -1 + 0 = -1

Formal charge of ammonium ion's atoms:

N = 5 - (0 + 4) = +1

H = 1 - (0 + 1) =   0

H = 1 - (0 + 1) =   0

H = 1 - (0 + 1) =   0

H = 1 - (0 + 1) =   0

net charge of ammonium ion = +1 + (4 x 0) = +1

Steps To Be Followed In Drawing A Lewis Electron Dot Structure

To summarize, the following steps may be followed to draw the Lewis electron dot structure of a chemical species.

1. Draw all possible structures by using single, double or triple pairs of electrons for atom-to-atom bonds.

2. Make sure that the structures contain the total valence electrons of all the atoms of the species.

3. If the species has a net charge, subtract electrons from the structure as indicated by the + charge or add electrons to the structure as indicated by the - charge.

4. Compute the formal charge of the atoms.

5. The structure that satisfies the octet rule for all atoms must be the correct Lewis structure.

Following the above steps, we're going to draw the electron dot structures of oxygen molecule (O₂) and cyanide ion (CN^-)

Drawing the Lewis Structure of a Molecule

Here are the 3 possible Lewis structures of N₂ molecule.

Structures I and II are incorrect because they violate the octet rule:

structure I has nitrogen atoms each containing only 6 electrons;

structure II has nitrogen atoms each containing only 5 electrons.

Structure III is the correct Lewis electron dot structure of N₂ molecule.

Drawing the Lewis Structure of an Ion

For the structures of cyanide ion as illustrated below, we added one electron to the total valence electrons of 9 (4 from C + 5 from N = 9) since it has a net charge of -1.

We rule out structures I and II because their atoms are electron deficient.

Structure III is the correct Lewis electron dot structure of cyanide ion.

The Resonance Theory

Sometimes, the step by step procedure we used above does not suffice because there are chemical species that have more than one possible Lewis electron dot structures.

One such species is the benzene molecule.

The Lewis structures shown below are equivalent but neither of them represents the actual structure of benzene molecule.

(We used single and double bonds here for easier viewing of the illustrations.)

Instead, benzene is represented by this structure:



This representation of a chemical species by a structure which is intermediate of 2 or more equivalent Lewis structures is called resonance.

The above structure is called resonance hybrid and structures I and II above are called resonance structures.

It should be noted here that resonance structures retain the same atomic arrangement; the only difference among these resonance structures is the arrangement of their electrons.

The Resonance Rules

There are several resonance rules that are used to determine the stability of resonance structures.

The most stable structures are the most important ones.

According to the resonance rules, the most stable structure is the one which has:

1. the greatest number of covalent bonds;

2. the least number of formal charges;

3. the - sign on the more electronegative atom and the + sign on the more electropositive atom, if formal charges are present.

We will use these rules to help us determine the most important Lewis structures of compounds having several resonance structures.

Determining The Most Stable Resonance Structure

Given below are the possible electron dot structures of CO₂.



We strike off structures I, II and III because of their electron deficient atoms.

Our choices then boil down to structures IV - VI.

We can see that structures V and VI are equivalent; we can use any of the two to compare with structure IV.

Being equivalent in the number of covalent bonds, we compare their number of formal charge.

Structure IV is more stable than structure V or VI because it has no formal charge. Hence, structure IV is the Lewis electron dot structure of CO₂.

Let's have another example: an ion this time.

The possible structures of thiocyanate ion (SCN^-) are given below.

How did we determine that C atom is the central atom?

We take the atom with the lowest valence electrons (except H) as the central atom because it has the maximum number of unpaired electrons that can be used to form covalent bonds with other atoms.

We exclude structures I - III from our choices because they don't follow the octet rule.

Since the above 3 structures have all equal number of covalent bonds, we will evaluate their stability based on the number of formal charge.

Structures IV and V are equivalent and are more stable than structure VI because they have less number of formal charge.

We choose structure IV as the Lewis structure of thiocyanate ion because the - sign is on the more electronegative atom.

Conclusion

By using the procedure and rules given above, you are now ready to draw the Lewis electron dot formula of most compounds having covalent bonds. Try the problems below.

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Problems

The best way to improve your problem solving skill in chemistry is to practice solving as many chemistry problems as possible. Here are some common ions and molecules. Draw their Lewis electron dot structures.

For more problems on other chemistry topics, go to www.tutorpartner.blogspot.com.

1.	C2H5NH3+
2.	CH3COO-
3.	H2O2
4.	HSO3-
5.	HCO3-
6.	CH3CN
7.	C2O4-2
8.	HNO2
9.	HCOCl
10.	N2
11.	CH3CONH2
12.	SO3-2
13.	C4H6
14.	NO3-
15.	NO2-

How to Derive a Gas Law From the Ideal Gas Law Equation

There are several equations that relate the properties of a gas. These relatioships of the pressure, the volume and the temperature for a fixed amount of a gas are given below.

The variables with subscript 1 indicate initial values; and those with subscript 2 indicate final values.

For one who is taking up chemistry for the first time, remembering all these equations is a little bit tricky.

Using the Ideal Gas Law equation, the above gas laws can be derived easily.

This way, you need to remember fewer equations.

The Ideal Gas Law Equation

PV = nRT

where:

• P is the pressure in atmosphere


• V is the volume in liters


• n is the amount of gas in moles


• R is the universal gas constant in L-atm/K-mol


• T is the temperature in Kelvin

This law relates the four variables of an ideal gas (pressure, volume, amount, and temperature) such that the value of PV/nT is always 0.082057.

It means that a change in the value of any of the four variables will bring about corresponding changes in the values of the other three variables so that the value of R remains constant.

Mathematically, it is equivalent to:

Deriving a Gas Law Equation

As mentioned earlier, we are dealing here with a fixed amount of gas.

It means the amount of gas remains the same from the initial state to the final state. So,

The value of nR must be constant.

Since the values of P₁V₁/T₁ and P₂V₂/T₂ are both equal to nR, then

Thus, we have derived the Combined Gas Law.

Using the above method, we can easily derive the other gas laws.

Suppose a problem states that the temperature must remain constant.

Using the Ideal Gas Law equation for the initial values and final values,

The value of nRT must be constant. Hence,

This is the Boyle's Law.

Now, you can derive the rest of the gas laws using the method we used here.

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Problems

Improve your problem solving skill in chemistry by solving as many chemistry problems as possible. Here are some problems dealing with gas laws. The answer is provided for each problem.

For more chemistry problems on gases and other chemistry topics, you can go to www.tutorpartner.blogspot.com.

Boyle's Law Problems

1.	Calculate the pressure required to reduce the volume of 18.0 L of gas at a pressure of 0.333 atm to 6.0 L with the temperature remaining constant.
	Answer: 0.999 atm
2.	Under constant temperature, 7000 mL of a sample gas was allowed to expand to a final volume of 23000 mL. If the initial pressure of the gas was 1085 torr, what was the final pressure of the gas after its expansion?
	Answer: 330 torr
3.	A sample of a gas which occupies 0.93 L has a pressure of 3 atm. Determine the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is maintained constant.
	Answer: 0.31 L
4.	Calculate the pressure required to reduce the volume of 3000 mL of gas at a pressure of 1260 torr to 1000 mL with the temperature remaining constant.
	Answer: 3780 torr
5.	Calculate the pressure required to reduce the volume of 24.0 L of gas at a pressure of 0.25 atm to 13.0 L with the temperature remaining constant.
	Answer: 0.462 atm
6.	At a pressure of 7 atm, a sample of gas has a volume of 8.634 L. If the pressure is reduced to 2.75 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 21.977 L
7.	Under constant temperature, 10000 mL of a sample gas was allowed to expand to a final volume of 17000 mL. If the initial pressure of the gas was 21489 torr, what was the final pressure of the gas after its expansion?
	Answer: 12641 torr
8.	At a pressure of 10 atm, a sample of gas has a volume of 0.497 L. If the pressure is reduced to 5 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 0.994 L
9.	A sample of a gas which occupies 2.8 L has a pressure of 1 atm. Determine the resulting volume of the gas if the pressure is changed to 8 atm and the temperature is maintained constant.
	Answer: 0.35 L
10.	Under constant temperature, 5.0 L of a sample gas was allowed to expand to a final volume of 19.0 L. If the initial pressure of the gas was 12.088 atm, what was the final pressure of the gas after its expansion?
	Answer: 3.181 atm
11.	Calculate the pressure required to reduce the volume of 14000 mL of gas at a pressure of 3281 torr to 3000 mL with the temperature remaining constant.
	Answer: 15311 torr
12.	At a pressure of 10.425 atm, a sample of gas has a volume of 0.214 L. If the pressure is reduced to 1.882 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 1.185 L
13.	Under constant temperature, 8.0 L of a sample gas was allowed to expand to a final volume of 25.0 L. If the initial pressure of the gas was 0.342 atm, what was the final pressure of the gas after its expansion?
	Answer: 0.109 atm
14.	At a pressure of 9.679 atm, a sample of gas has a volume of 0.397 L. If the pressure is reduced to 5.373 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 0.715 L
15.	At a pressure of 8.537 atm, a sample of gas has a volume of 0.45 L. If the pressure is reduced to 2.531 atm, what will be the resulting volume of the gas? Temperature is maintained constant.
	Answer: 1.518 L

Charles' Law Problems

1.	Calculate the final volume of a 3.65 L gas at 92° C when the temperature is brought down to 9° C as the pressure remains unchanged.
	Answer: 2.82 L
2.	Calculate the final volume of a 7.39 L gas at 43° C when the temperature is brought down to 13° C as the pressure remains unchanged.
	Answer: 6.69 L
3.	A sample of gas with a volume of 6.0 L at -17° C was allowed to expand to 21.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: 623° C
4.	18 L of gas at 294° C was compressed to 2 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -210° C
5.	Maintaining a constant pressure, a sample of gas having a volume of 5.0 L at -127° C is heated to 369° C. What is the final volume of the gas?
	Answer: 22.0 L
6.	20 L of gas at 582° C was compressed to 4 L at constant pressure. Determine the resulting temperature of the gas.
	Answer: -102° C
7.	A sample of gas with a volume of 3.0 L at -145° C was allowed to expand to 17.0 L while maintaining a constant pressure. What was the new temperature of the gas?
	Answer: 452° C
8.	Maintaining a constant pressure, a sample of gas having a volume of 9.0 L at -210° C is heated to -7° C. What is the final volume of the gas?
	Answer: 38.0 L

Gay-Lussac's Law Problems

1.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.795 atm at 9° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 88° C?
	Answer: 1.043 atm
2.	A gas cylinder containing a certain gas is found to have a pressure of 18.053 atm at 57° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 0° C, calculate the resulting pressure inside the cylinder.
	Answer: 18.053 atm
3.	A sample of a gas has a pressure of 10 atm and a temperature of 32° C. Find the corresponding change in the temperature if the pressure is (a) reduced to 5 atm, (b) tripled. Assume the volume of the gas is constant.
	Answer: (a) -120° C, (b) 642° C
4.	A sample of a gas sealed in a vessel was determined to have a pressure of 1.007 atm at 2° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 81° C?
	Answer: 1.304 atm
5.	A gas cylinder containing a certain gas is found to have a pressure of 147.177 atm at 50° C. If the cylinder is immersed in a cooling solution so that the temperature is lowered by 1° C, calculate the resulting pressure inside the cylinder.
	Answer: 146.721 atm
6.	A sample of a gas sealed in a vessel was determined to have a pressure of 0.906 atm at 3° C. What would be the pressure of the gas if the vessel was heated to increase the temperature by 84° C?
	Answer: 1.182 atm

Combined Gas Law Problems

1.	What will be the final temperature of a 51.413 L gas at 27° C if its volume is changed to 18.283 L and its pressure changed from 0.974 atm to 4.063 atm?
	Answer: 172° C
2.	784.96 L of a sample of gas has a pressure of 0.23 atm and a temperature of 19° C. After compressing the volume down to 1.37 L and increasing the temperature to 98° C, calculate the change in the pressure of the gas.
	Answer: 167.435 atm
3.	69.73 L of a sample of gas has a pressure of 0.98 atm and a temperature of 29° C. After compressing the volume down to 7.47 L and increasing the temperature to 72° C, calculate the change in the pressure of the gas.
	Answer: 10.451 atm
4.	An expandable air balloon containing 88.2 L of a sample of a gas has a pressure of 1.039 atm at 47° C. What will be the resulting volume of the gas if the pressure is changed to 7 atm and the temperature is increased by 15° C?
	Answer: 13.7 L
5.	What will be the final temperature of a 161.519 L gas at 12° C if its volume is changed to 24.129 L and its pressure changed from 0.225 atm to 11.314 atm?
	Answer: 1866° C
6.	An expandable air balloon containing 8.3 L of a sample of a gas has a pressure of 31.941 atm at 42° C. What will be the resulting volume of the gas if the pressure is changed to 9 atm and the temperature is increased by 14° C?
	Answer: 30.8 L

Ideal Gas Law Problems

1.	A 29.135 L tank containing a sample of gas has a pressure of 0.968 atm and a temperature of 70° C. Find the moles of gas contained inside the tank.
	Answer: 1.002 mol
2.	Calculate the volume of 9.385 mol of a gas confined in a sealed vessel with a pressure of 39.66 atm and a temperature of 36° C.
	Answer: 6 L
3.	Given a volume of 25.07 L of the following gases at STP : (a) SO2, (b) N2, (c) H2S, find the weight in grams of each gas. (mol. wt.: SO2=64.0588, N2=28.0134, H2S=34.0758)
	Answer: (a) 71.69 g, (b) 31.35 g, (c) 38.14 g
4.	Given a weight of 8.8 g of each of the following gases: (a) NH3, (b) He, (c) NH3, calculate the volume of each gas at STP. (mol. wt.: NH3=17.0304, He=4.0026, NH3=17.0304)
	Answer: (a) 11.57 L, (b) 49.25 L, (c) 11.57 L

A Note on Dilution and Concentrations of Solutions

Problems dealing with dilution of solutions from one concentration to another concentration can sometimes be confusing.

In order to have a better understanding of it, we will use graphical representation of the solute-solvent components of a solution to help us visualize the problem.

A typical example of this kind of problem is given below:

Given 2 L of 2 M NaCl solution, make a solution having a concentration of 1 M NaCl solution. What must be the final volume of the solution?

We will use a graphical representation like the one below to explain this problem.

The blue box represents one liter of solution, and each white ellipse represents one mole of solute ( NaCl ).

The blue box containing the two ellipses represents the concentration of the solution ( 2 M NaCl = 2 moles NaCl/1 liter of solution ).

So our given solution of 2 L of 2 M NaCl will be represented like this:

By inspection, we can see that we have a total of 4 moles NaCl in 2 L of solution.

Mathematically, finding the total amount of solute in a solution is equivalent to:

Volume of solution (V)  x  Concentration of solution (C)  =  amount of solute

or simply,

V x C  =  amount of solute                         (1)

Since the problem requires us to make 1 M NaCl solution from the given solution by diluting it (adding water to the solution), the total amount of solute must remain the same before and after dilution.

To account for the 4 moles of NaCl, the graphical representation for our final solution will be like this:

By inspection again, we can see that the volume of the final solution is now increased to 4 liters.

Using equation number 1, we can find the final volume of the solution mathematically by solving for V:

So the problem can be represented like this:

2 L of 2 M NaCl

4 L of 1 M NaCl

amount of solute before dilution

=

amount of solute after dilution

Substituting equation 1 into this, and using subscript 1 and 2 to distinguish the initial solution from the final solution, gives us:

V1 x C1

=

V2 x C2 (2)

Equation 2 is the formula for dilution problems involving different kinds of concentrations.









Solving for V₂:

Just knowing that the amount of solute does not change and employing the dimensional analysis to the problem will enable you to figure out the formula by yourself, as long as you know the definitions of different concentrations.

Here are the commonly used concentrations in chemistry problems and their definitions:

The last two expressions are used in very dilute solutions. They are used to express trace amount of solute in aqueous solutions. (1 gram = 1,000 milligrams = 1,000,000 micrograms)

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Problems

The best way to improve your problem solving skill in chemistry is to solve as many chemistry problems as you can find. Here are some problems on dilutions and concentrations of solutions. The answer is provided for each problem.

Dilution Problems

1.	What is the volume of the diluted solution of 10.0 mL of 4.34 M K2HPO4 if its final concentration is 0.55 M K2HPO4?
	Answer: 78.9 mL
2.	If 15.0 mL of 1.9 M Cu(NO3)2 solution is diluted to 45.0 mL, calculate the new molarity of the solution.
	Answer: 0.6 M
3.	23.6 mL of 1.07 M K2SO4 solution was prepared from an unknown volume of 4.21 M K2SO4 solution. What was the initial volume of the solution?
	Answer: 6.0 mL
4.	20.00 mL of an unknown concentration of LiNO3 solution was diluted to 64 mL to make a concentration of 0.59 M LiNO3. Determine the initial concentration of the diluted solution.
	Answer: 1.89 M
5.	22.00 mL of an unknown concentration of NaClO3 solution was diluted to 63 mL to make a concentration of 0.5 M NaClO3. Determine the initial concentration of the diluted solution.
	Answer: 1.43 M
6.	If 8.0 mL of 0.8 M NiCl2 solution is diluted to 63.0 mL, calculate the new molarity of the solution.
	Answer: 0.1 M
7.	20.4 mL of 1.44 M NH4HCO3 solution was prepared from an unknown volume of 1.63 M NH4HCO3 solution. What was the initial volume of the solution?
	Answer: 18.0 mL
8.	21.00 mL of an unknown concentration of KCH3COO solution was diluted to 64 mL to make a concentration of 0.3 M KCH3COO. Determine the initial concentration of the diluted solution.
	Answer: 0.9 M
9.	What is the volume of the diluted solution of 17.0 mL of 4.06 M NaCl if its final concentration is 2.84 M NaCl?
	Answer: 24.3 mL
10.	If 24.0 mL of 3.9 M K2Cr2O7 solution is diluted to 38.0 mL, calculate the new molarity of the solution.
	Answer: 2.5 M

Concentration Problems

1.	Determine the weight in grams of the solute contained in a given volume and a given concentration of the following solutions: (a) 3.0 L of 1.97 M Ca(NO3)2.4H2O solution, (b) 14.0 mL of 1.84 M CuS solution, (c) 23.0 mL of 0.91 M Cs2SO4 solution. (mol. wt.: Ca(NO3)2.4H2O=236.1506, CuS=95.606, Cs2SO4=361.8684)
	Answer: (a) 1395.7 g, (b) 2.5 g, (c) 7.6 g
2.	Calculate the final molarity (M) of a mixed solution of 15.0 mL of 4.65 M NaCH3COO solution and 10.0 mL of 2.96 M NaCH3COO solution, assuming their volumes are additive.
	Answer: 4 M
3.	Calculate the resulting molarity (M) of NO3-1 from the mixture of 17.00 mL of 1.42 M NaNO3 solution and 18.00 mL of 4.2 M Mg(NO3)2 solution, assuming their volumes are additive.
	Answer: 5.01 M
4.	Determine the weight in grams of the solute contained in the following solutions: (a) 14.0 mL of 0.78 M KOH solution, (b) 4.0 L of 1.81 M Ca(CH3COO)2 solution, (c) 6.0 L of 3.03 M AgNO3 solution. (mol. wt.: KOH=56.1056, Ca(CH3COO)2=158.169, AgNO3=169.8729)
	Answer: (a) 0.6 g, (b) 1145.1 g, (c) 3088.3 g
5.	What is the molarity (M) of the following solutions: (a) 96% (w/w) glacial acetic acid solution with a density of 1.06 g/mL, (b) 35% (w/w) ammonium hydroxide solution with a density of 0.88 g/mL. (mol. wt.: CH3COOH=60.0524, NH4OH=35.0456)
	Answer: (a) 16.95 M, (b) 8.79 M
6.	Calculate the resulting molarity (M) of Cl-1 from the mixture of 10.00 mL of 2.25 M CoCl3 solution and 20.00 mL of 3.2 M MgCl2 solution, assuming their volumes are additive.
	Answer: 6.52 M
7.	Calculate the weight of the solute required in order to prepare the following solutions: (a) 18.0 mL of 5.21 M Na4P2O7 solution, (b) 2.0 L of 1.7 M NH4H2PO4 solution, (c) 5.0 L of 1.82 M Al2(SO4)3 solution. (mol. wt.: Na4P2O7=265.9024, NH4H2PO4=115.02546, Al2(SO4)3=342.13588)
	Answer: (a) 24.9 g, (b) 391.1 g, (c) 3113.4 g
8.	What is the molarity (M) of the following solutions: (a) 40% (w/w) hydrobromic acid solution with a density of 1.38 g/mL, (b) 97% (w/w) concentrated sulfuric acid solution with a density of 1.84 g/mL. (mol. wt.: HBr=80.9119, H2SO4=98.0734)
	Answer: (a) 6.82 M, (b) 18.2 M
9.	Determine the weight in grams of the solute contained in the following solutions: (a) 9.0 mL of 1.4 M NaClO3 solution, (b) 1.0 L of 4.24 M Zn(NO3)2.6H2O solution, (c) 4.0 L of 3.31 M Ca(NO3)2.4H2O solution. (mol. wt.: NaClO3=106.44097, Zn(NO3)2.6H2O=297.481, Ca(NO3)2.4H2O=236.1506)
	Answer: (a) 1.3 g, (b) 1261.3 g, (c) 3126.6 g
10.	What is the molarity of a 23.0 mL solution if it contains: (a) 6.23 g of CdCO3, (b) 1.78 g of K2S2O5, (c) 7.47 g of Na2HPO4.2H2O? (mol. wt.: CdCO3=172.4192, K2S2O5=222.3136, Na2HPO4.2H2O=177.9892)
	Answer: (a) 1.6 M, (b) 0.3 M, (c) 1.8 M