The solubility of barium hydroxide in water at 20 degrees Celsius is 1.85g/100g water. A solution is made up of 256 mg in 35.0 g of water at 20 degrees Celsius. Is the solution saturated?

(35.0 g H2O)/1 x (1.85 g Ba(OH)2)/(100 g H2O) x (1000 mg Ba(OH)2)/(1 g Ba(OH)2) = 648 mg Ba(OH)2 dissolved in 35.0 g H2O to make a saturated solution.

Since only 256 mg Ba(OH)2 is dissolved in 35.0 g H2O, the solution is unsaturated.

If you want to lower water’s freezing point 15 degrees celsius by adding NaCl, what must the molality of the salt solution be (Kf for water is 1.86 degrees celsius per molal)?

15 oC = (2 particles)(1.86 oC/m)(molal NaCl (aq))
molal NaCl(aq) = 4.03 molal

What is the freezing point of a solution made by mixing 30.0 grams of NH3 with 300.0 grams of water (Kf for water is 1.86 degrees celsius per molal)?

1. Solve for the Molality of NH3(aq)

Molality (molal) is in units of moles of solute per kilograms of solvent. Ammonia is the solute and water is the solvent.

molal NH3(aq) = (moles NH3)/(kg H2O)

1a. Calculate Moles Solute (NH3)

moles NH3 = (30.0 g NH3)/1 x (1 mol NH3)/(17.04 g NH3) = 1.76 mol NH3

1b. Calculate Kilograms Solvent (H2O)

kg H2O = (300.0 g H2O)/1 x (1 kg H2O)/(1000 g H2O) = 0.3000 kg H2O

1c. Calculate Molality NH3(aq)

molal NH3(aq) = (1.76 mol NH3)/(0.3000 kg H2O) = 5.87 mol NH3/kg H2O = 5.87 molal NH3(aq)

2. Solve for the Freezing Point of the Solution [NH3(aq)]

2a. Calculate the Decrease in Freezing Point

The decrease in freezing point is equal to the van’t Hoff factor multiplied by the molal freezing-point depression constant for water and the molality of ammonia solute particles in water.

Decrease in freezing point = (1)(1.86 oC/molal)(5.87 molal) = 10.9 oC

2b. Calculate the Freezing Point of the Solution

The freezing point of the solution is equal to the difference between the normal freezing point of pure water and the decrease in freezing point value.

Freezing point of the solution = (0 oC) – (10.9 oC) = -11 oC

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (e) 35.5 mL, (f) 50.0 mL.

The initial pH of 35.0 mL 0.150 M CH3COOH is 2.784. After 17.5 mL NaOH added, the pH is 4.74. After 34.5 mL NaOH added, the pH is 6.584. After 35.0 mL NaOH added, the pH is 8.810. After 35.5 mL NaOH added, the pH is 9.88. After 50.0 mL NaOH added, the pH is 11.35.

Weak Acid Equilibrium Calculations

Acetic acid has a Ka value of 1.8E-5. This will be very important to use when calculating the pH of the initial solution as well as the solutions after the specified amounts of base have been added.

(a) Initial pH of 35.0mL 0.150 M CH3COOH

To calculate the initial pH of the weak acid solution (when 0 mL NaOH have been added), the initial concentration of acetic acid and the acid-ionization constant is used to find the amount of hydrogen ions at equilibrium.

[H+][CH3COO-] =Ka [CH3COOH]

CH3COOH <–> CH3COO- + H+

(x)(x)/(0.150-x) =1.8E-5

assume x << 5%

(x^2)/(0.150) =1.8E-5

x=[H+]=1.64E-3 M

CH3COOH <–> CH3COO- + H+
i0.150 M0 M0 M
c-x+x+x
e0.150 – x x x

To make sure the assumption of percent ionization is less than 5% we divide the change, x, by the initial molarity of weak acid, then multiply the value by 100:

assumption for 5% rule of ionization to eliminate the need to use the quadratic formula

The assumption is valid meaning the change in x value calculated can be used to find the pH of the initial 35.0 mL 0.150 M acetic acid solution:

x=[H+]=1.64E-3 M

pH = -log(1.64E-3) = 2.784

Mole Calculations for the Titration

It is important to first calculate how many moles of acetic acid are present before any additions of sodium hydroxide:

Converting 35.0 mL acetic acid of 0.150 M acetic acid to moles of acetic acid

0.00525 moles of acetic acid are initially present in the 35.0 mL 0.150 M acetic acid solution.

pH Calculations Before the Equivalence Point

(b) pH after 17.5 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

Acetic acid reacting with hydroxide ions in the buffer region of titration before the equivalence point.

CH3COOH + OH- —-> CH3COO- + H2O

There are 0.00525 moles of acetic acid initially present in 35.0 mL 0.150 M acetic acid solution. When 17.5 mL 0.150 M NaOH are added, 0.002625 moles of hydroxide ions are added to the weak acid solution and react with acetic acid:

(moles)CH3COOH–>CH3COO-+H2O
before addition0.00525 mol0 molx
addition OH--0.002625 mol+0.002625 molx
after addition0.002625 mol0.002625 mol

After addition of 17.5 mL 0.150 M NaOH, all hydroxide ions have reacted to produce 0.002625 moles of acetate anion, leaving 0.002625 moles of acetic acid unreacted in solution. The new volume of the solution will equal the sum of the initial volume of weak acid and the volume of sodium hydroxide added. With a new total volume, the concentration values of acetic acid and acetate anion will equal the moles after addition divided by the new total volume of solution.

Total Volume = 35.0 mL + 17.5 mL = 52.5 mL solution

[CH3COOH] = [CH3COO-] = 0.002625 mol/0.0525 L = 0.05 M

[H+][CH3COO-] =Ka [CH3COOH]

CH3COOH <–> CH3COO- + H+

(x)(0.05+x)/(0.05-x) =1.8E-5

assume x << 5%

(0.05x)/(0.05) =1.8E-5

x=[H+]=1.8E-5 M

CH3COOH <–> CH3COO- + H+
i0.05 M0.05 M0 M
c-x+x+x
e0.05 – x 0.05 + x x

1.8E-5/0.05 x 100 << 5% meaning the assumption is valid.

x=[H+]=1.8E-5 M

pH = -log(1.8E-5) = 4.74

(c) pH after 34.5 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

Acetic acid reacting with hydroxide ions before the equivalence point.

There are 0.00525 moles of acetic acid initially present in 35.0 mL 0.150 M acetic acid solution. When 34.5 mL 0.150 M NaOH are added, 0.005175 moles of hydroxide ions are added to the weak acid solution and react with acetic acid:

(moles)CH3COOH–>CH3COO-+H2O
before addition0.00525 mol0 molx
addition OH--0.005175 mol+0.005175 molx
after addition7.5E-5 mol0.005175 mol

After addition of 34.5 mL 0.150 M NaOH, all hydroxide ions have reacted to produce 0.005175 moles of acetate anion, leaving 7.5E-5 moles of acetic acid unreacted in solution. The new volume of the solution will equal the sum of the initial volume of weak acid and the volume of sodium hydroxide added. With a new total volume, the concentration values of acetic acid and acetate anion will equal the moles after addition divided by the new total volume of solution.

Total Volume = 35.0 mL + 34.5 mL = 69.5 mL solution

[CH3COO-] = 0.005175 mol/0.0695 L = 0.0745 M

[CH3COOH] = 7.5E-5 mol/0.0695 L = 1.08E-3 M

[H+][CH3COO-] =Ka [CH3COOH]

CH3COOH <–> CH3COO- + H+

(x)(0.0745 +x)/(1.08E-3-x) =1.8E-5

assume x << 5%

(0.0745x)/(1.08E-3) =1.8E-5

x=[H+]=2.61E-7 M

CH3COOH <–> CH3COO- + H+
i1.08E-3 M0.0745 M0 M
c-x+x+x
e1.08E-3 – x 0.0745 + x x

2.61E-7/1.08E-3 x 100 << 5% meaning the assumption is valid.

x=[H+]=2.61E-7 M

pH = -log(2.61E-7) = 6.584

pH Calculations at the Equivalence Point

(d) pH after 35.0 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

Acetic acid reacting with hydroxide ions at the equivalence point.

There are 0.00525 moles of acetic acid initially present in 35.0 mL 0.150 M acetic acid solution. When 35.0 mL 0.150 M NaOH are added, 0.00525 moles of hydroxide ions are added to the weak acid solution and react completely with all moles of acetic acid present:

(moles)CH3COOH–>CH3COO-+H2O
before addition0.00525 mol0 molx
addition OH--0.00525 mol+0.00525 molx
after addition0 mol0.00525 mol

After addition of 35.0 mL 0.150 M NaOH, all hydroxide ions have reacted to produce 0.00525 moles of acetate anion, leaving 0 moles of acetic acid unreacted in solution. The new volume of the solution will equal the sum of the initial volume of weak acid and the volume of sodium hydroxide added. With a new total volume, the concentration value for acetate anion will equal the moles after addition divided by the new total volume of solution.

Total Volume = 35.0 mL + 35.0 mL = 70.0 mL solution

[CH3COO-] = 0.00525 mol/0.0700 L = 0.0750 M

CH3COO- + H2O <–> CH3COOH + OH

[OH-][CH3COOH] =Kb [CH3COO-]

(Ka)(Kb) = 1E-14

Kb = (1E-14)/ (1.8E-5) = 5.6E-10

(x)(x)/(0.0750-x) =5.6E-10

assume x << 5%

(x^2)/(0.0750) =5.6E-10

x=[OH-]=6.45E-6 M

CH3COO- <–> CH3COOH +OH-
i0.0750 M0 M0 M
c-x+x+x
e0.0750 – x x x

6.45E-6/0.0750 x 100 << 5% meaning the assumption is valid.

x=[OH-]=6.45E-6 M

pOH = -log(6.45E-6) = 5.190

pH = 14 – pOH = 14 – 5.190 = 8.810

pH Calculations After the Equivalence Point

(e) pH after 35.5 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

After the equivalence point, any additional volumes of sodium hydroxide added will be solely used to calculate the pH of the solution. Since 35.0 mL NaOH 0.150 M NaOH was needed to react completely with 35.0 mL 0.150 M CH3COOH, the volume used to calculate the moles of hydroxide ions present will be the difference between the volume of sodium hydroxide added and the volume of base at the equivalence point.

Excess Volume NaOH = 35.5 mL – 35.0 mL = 0.5 mL excess 0.150 M NaOH

pOH = -log(7.5E-5) = 4.12

pH = 14 – 4.12 = 9.88

(f) pH after 50.0 mL 0.150 M NaOH added to 35.0mL 0.150 M CH3COOH solution

After the equivalence point, any additional volumes of sodium hydroxide added will be solely used to calculate the pH of the solution. Since 50.0 mL NaOH 0.150 M NaOH was needed to react completely with 35.0 mL 0.150 M CH3COOH, the volume used to calculate the moles of hydroxide ions present will be the difference between the volume of sodium hydroxide added and the volume of base at the equivalence point.

Excess Volume NaOH = 50.0 mL – 35.0 mL = 15.0 mL excess 0.150 M NaOH

pOH = -log(2.25E-3) = 2.648

pH = 14 – 2.648 = 11.35

Household hydrogen peroxide is an aqueous solution containing 3.0% hydrogen peroxide by mass. What is the molarity of this solution? (Assume a density of 1.01 g/mL.)

1. Calculate Moles H2O2 in Solution

3.0% H2O2 by mass = (grams H2O2)/ (grams solution) x 100

assume 100 grams of solution

3.0% H2O2 by mass = (grams H2O2)/ (100 g solution) x 100
0.030 = (grams H2O2)/(100 g solution)
3.0 = grams H2O2
(3.0 grams H2O2)/1 x (1 mole H2O2)/(34.02 g H2O2) = 0.088 moles H2O2

2. Calculate Liters of Solution

(100 g sol’n)/1 x (1 mL sol’n)/(1.01 g sol’n) x (1 L sol’n)/(1000 mL sol’n) = 0.0990 L sol’n

3. Calculate Molarity H2O2(aq)

Molarity H2O2 = (mol H2O2)/(L solution) = (0.088 mol H2O2)/(0.099 L sol’n) = 0.89 M

A solution of formic acid (HCOOH, Ka = 1.8 X 10^-4) has a pH of 2.70. Calculate the initial concentration of formic acid in this solution.

Since the acid given, formic acid, has an acid-ionization constant, Ka, we can conclude formic acid is a weak acid that only partially dissociates or does not ionize completely.

For weak acid problems, formic acid’s balanced chemical equation showing the dissociation of the acid into its ions, the acidic proton and its conjugate base, should be written alongside the Ka expression.

HCOOH <—> H+ + HCOO-

Ka = [H+][HCOO-]/
[HCOOH]

Equilibrium concentration values in units of moles per liter are used in Ka expressions.

[H+][HCOO-]/ = 1.8×10^-4 [HCOOH]

(M H+)(M HCOO-)/ = 1.8×10^-4 (M HCOOH)

The pH value for the formic acid solution is given. After a weak acid solution reaches equilibrium, the final [H+] concentration is used to calculate the pH value of the solution. Since the problem has given us the final pH value of the solution after equilibration, the equilibrium value for the concentration of hydrogen ions can be calculated and we will work backwards to fill in the rest of the ICE table:

Equilibrium [H+]

[H+]=10^-pH

[H+]=10^-2.70

[H+]= 2.0×10^-3 M

HCOOH <—> H+ + HCOO-
i
c
e2.0E-3

Change in Concentration

initial [H+]= 0

equilibrium [H+]= 2.0×10^-3 M

change in concentration, x = equilibrium [H+] – initial [H+] = 2.0×10^-3 M – 0 M = 2.0×10^-3 M

(2.0E-3)^2/(initial – 2.0E-3) = 1.8×10^-4

HCOOH <—> H+ + HCOO-
i????0 0
c-x+x+x
e2.0E-3
HCOOH <—> H+ + HCOO-
i????0 0
c2.0E-3 +2.0E-3 +2.0E-3
einitial – 2.0E-32.0E-32.0E-3

(initial – 2.0E-3)(1.8E-4) = (2.0E-3)^2

initial – 2.0E-3 = 0.0221

initial [HCOOH] = 0.0241 M

The pH of 1.00 x 10^-2 M solution of cyanic acid (HOCN) is 2.77 at 25 degrees Celsius. Calculate Ka for HOCN from this result.

To calculate the acid-ionization constant for cyanic acid, HOCN, the balanced chemical equation showing the ionization of cyanic acid is needed so we can then write the acid-ionization expression to eventually solve for Ka, the acid-ionization constant.

HOCN (aq) <—> H+ (aq) + OCN- (aq)

Ka = [H+][OCN-] / [HOCN]

To begin to piece together the equilibrium values needed for the Ka expression, an ICE table is used to map out the information given. An initial solution of 1.00 E-2 M HOCN is given:

HOCN<–>H+OCN-
initial1.00E-2 M0 M0 M
change-x+x+x
equilibrium

The pH of acid-base reactions is measured after a solution has reached equilibrium so we can use the pH value of 2.77 given to calculate the equilibrium concentration of hydrogen ions:

10^-pH = [H+] therefore, 10^-2.77 = [H+] = 1.698 E-3 M

HOCN<–>H+OCN-
initial1.00E-2 M0 M0 M
change-x+x+x
equilibrium1.698E-3 M

The initial concentration of H+ is 0 M and since we now know the equilibrium concentration of H+, we can solve for the change variable, x, needed to fill in the rest of the values in the ICE table:

0 + x = 10^-2.77 therefore, x = 10^-2.77 = 1.698E-3 M

HOCN<–>H+OCN-
initial1.00E-2 M0 M0 M
change– 1.698E-3 M + 1.698E-3 M + 1.698E-3 M
equilibrium8.30E-3 M 1.698E-3 M 1.698E-3 M

Now, that the equilibrium values are known for cyanate, OCN-, hydrogen ion, H+, and cyanic acid, HOCN, the acid-ionization equilibrium expression can be used to calculate the value of Ka, the acid-ionization constant.

HOCN (aq) <—> H+ (aq) + OCN- (aq)

[H+] = [OCN-] = 1.698E-3 M

[HOCN] = 8.30E-3 M

Ka = [H+][OCN-]/ [HOCN]

Ka= ( 1.698E-3 M)^2/ (8.30E-3 M)

HOCN Ka = 3.47E-4

Titration Curve for 5.00 mL 0.010 M HCl (aq) with 0.010 M Ca(OH)2 (aq)

(a) Sketch the titration curve for the titration of 5.00 mL 0.010 M HCl (aq) with 0.010 M Ca(OH)2 (aq) showing the pH of the initial and final solutions and the pH at the stoichiometric point. What volume of titrant has been added at (b) the stoichiometric point; (c) the halfway point of the titration?

Atkins, P., & Jones, L. (2010). Chemical Principles: The Quest for Insight with Unique International Edition Problem Sets, Fifth Edition. New York, NY: W. H. Freeman and Company.

Initial pH of 5.00 mL 0.010 M HCl (aq)

Strong acids experience 100% ionization so the concentration of hydrochloric acid will have a one-to-one ratio to its products, hydrogen ion (proton) and chloride ion.

HCl (aq) –> H+ (aq) + Cl- (aq)

[HCl] = [H+] therefore, 0.010 M HCl = 0.010 M H+

pH = -log([H+]) therefore, pH= -log(0.010 M)

The initial pH of 5.00 mL 0.010 M HCl (aq) is equal to 2. The titration curve will have an x-axis labeled mL of Ca(OH)2 added and a y-axis labeled pH. The first coordinate will be (0.00 mL base added, pH 2).

Volume of 0.010 M Ca(OH)2 at the Stoichiometric Point with 5.00 mL 0.010 HCl

For any titration, moles of acid is equal to moles of base at the stoichiometric point.

5.00 mL HCl1 L HCl0.010 mol HCl1 mol H+1 mol OH-1 mol Ca(OH)21 L Ca(OH)21000 mL Ca(OH)2
11000 mL HCl1 L HCl1 mol HCl1 mol H+2 mol OH-0.010 mol Ca(OH)21 L Ca(OH)2

Using the conversion table above from mL HCl to mL Ca(OH)2, 2.5 mL of 0.010 M Ca(OH)2 is needed to reach the stoichiometric point wheres moles of acid will equal moles of base.

An alternative method can be used to solve for the volume of calcium hydroxide at the stoichiometric point with the equation MaVa = MbVb, where concentration of acid times volume of acid is equal to concentration of base times volume of base.

It is important to note that calcium hydroxide contains 2 moles of hydroxide ions, therefore the molarity of base will be 0.02 M to use in the equation:

(0.01 M acid)(5.00 mL acid) = (0.02 M base)(Vb), where Vb = 2.5 mL base

Stoichiometric pH of 5.00 mL 0.010 M HCl (aq) and 2.5 mL 0.010 M Ca(OH)2

Strong-acid-strong-base neutralization reactions will have a pH equal to 7 at the stoichiometric point. All moles of hydrogen ion have reacted with all moles of hydroxide ions to give the net ionic equation:

H+(aq) + OH-(aq) –> H2O(l)

H+(aq OH-(aq) –> H2O(l)
initial moles5 E-5 mol0x0
change: added OH--5E-5 molall moles react with H+x+5E-5 mol
final moles0 mol0 molx5E-5 mol

Since the only remaining compound in solution is H2O(l), The autoionization constant of water, Kw, can help to confirm the pH of the solution at the stoichiometric point:

Kw=1E-14=[OH-][H+], where x=[OH-]=[H+]

1E-14 = x^2, where x= 1E-7, therefore [H+]= 1E-7, and pH= -log(1E-7)

The pH of the neutralization reaction at the stoichiometric point will be equal to 7, and is a neutral solution.

The coordinate for the stoichiometric point on the graph will be (2.5 mL base added, pH 7).

Volume of 0.010 M Ca(OH)2 added at the halfway point of the titration of 5.00 mL 0.010 M HCl

Since we know 2.5 mL of calcium hydroxide is needed to reach the stoichiometric point, this volume of base is halved to reach the halfway point of the titration, therefore 1.25 mL of 0.010 M Ca(OH)2 is added at the halfway point of titration of 5.00 mL 0.010 M HCl.

The halfway point of the titration is also where half the moles of acid has reacted with base and half the moles of the acid still remains unreacted in the solution.

H+(aq) + OH-(aq) –> H2O(l)

0.010 M Ca(OH)2= 0.020 M OH-, therefore, 0.020 mol OH- is added per every L of solution and 1.25E-3 L solution will add 2.5E-5 mol OH- to the reaction at the halfway point of titration:

H+(aq OH-(aq) –> H2O(l)
initial moles5 E-5 mol0x0
change: added OH--2.5E-5 molall moles react with H+x+2.5E-5 mol
final moles2.5E-5 mol0 molx2.5E-5 mol

The pH of the solution at the halfway point of titration will depend solely on the leftover amount of hydrogen ions in solution and the total volume present after the addition of 1.25 mL base.

The concentration of hydrogen ions at the halfway point is equal to 2.5E-5 mol H+ divided by 6.25E-3L total volume of solution:

[H+] = 2.5E-5 mol/6.25E-3 L = 0.004 M H+, therefore pH = – log(0.004M) = 2.40

The coordinate for halfway point of titration to sketch on the graph will be (1.25 mL base added, pH 2.40)

pH of the Final Solution for the Titration of 5.00 mL 0.010 M HCl with 0.010 M Ca(OH)2

To complete the sketch of the titration graph, an accurate representation of the pH after the stoichiometric point needs to be shown. Once we have reached the stoichiometric point for a strong-acid-strong-base neutralization reaction, the pH is calculated based solely on the concentration of hydroxide ions added in excess. These hydroxide ions are added “in excess” because all moles of acid have reacted. Any volume of base that is greater than the amount needed to reach the stoichiometric point is effective for this calculation: 3.75 mL and 5 mL of Ca(OH)2 will be used.

5E-5 moles OH- ions will react with moles of H+ ions to reach the stoichiometric point.

3.75E-3 L base with a concentration of 0.020 M OH- ions will yield a total of 7.5E-5 moles OH-. There will be 2.5E-5 moles OH- in excess with a total solution volume of 8.75 mL. The concentration of hydroxide ions will be equal to 2.5E-5 mol/ 8.75E-3 L= 0.0025 M OH-.

The pOH of the solution will be equal to -log(0.0025 M)= 2.60 , therefore the pH will be equal to 14-pOH = 11.40 when 3.75 mL of base is added.

On the sketch of the titration graph, a coordinate will be (3.75 mL base added, pH 11.40).

5E-3 L base with a concentration of 0.020 M OH- ions will yield a total of 1E-4 moles OH-. There will be 5E-5 moles OH- in excess with a total solution volume of 10 mL. The concentration of hydroxide ions will be equal to 5E-5 mol/10E-3 L = 0.005 M OH-.

The pOH of the solution will be equal to -log(0.005 M) = 2.30, therefore the pH will be equal to 14-pOH = 11.70 when 5.00 mL of base is added.

On the sketch of the titration graph, a coordinate will be (5.00 mL base added, pH 11.70).

Sketch a Curve of the Titration of 5.00 mL 0.010 M HCl with 0.010 M Ca(OH)2

mL base addedpH
02
1.252.4
2.57
3.7511.4
511.7

Metric Prefixes

prefix symbol multiplier to base unitex:equivalence to base unit (meters)
gigaG 1,000,000,000 or 1E9 1 Gm = 1E9 m
or
1 Gm = 1,000,000,000 m
megaM1,000,000 or 1E61 Mm = 1E6 m
or
1 Mm = 1,000,000 m
kilok1,000 or 1E31 km = 1E3 m
or
1 km = 1,000 m
dekada10 or 1E11 dam = 1E1 m
or
1 dam = 10 m
decid0.1 or 1E-11 dm = 1E-1 m
or
1 dm = 0.1 m
centic0.01 or 1E-21 cm = 1E-2 m
or
1 cm = 0.01 m
millim0.001 or 1E-31 mm = 1E-3 m
or
1 mm = 0.001 m
microu0.000001 or 1E-61 um = 1E-6 m
or
1 um = 0.000001 m
nanon0.000000001 or 1E-91 nm = 1E-9 m
or
1 nm = 0.000000001 m
picop0.000000000001 or 1E-12 1 pm = 1E-12 m
or
1 pm = 0.000000000001 m