Question

1.

Calculate the approximate pH of the following aqueous solutions at 25C. (Note: consider one decimal place for the pH answer.)

a) H₃PO₄ (0.20 M), (pK_a1=2.15, pK_a2=7.15, pK_a3=12.38)

b) Na₂CO₃ (0.20 M), (pK_a H₂CO₃/HCO₃^? = 6.35, pKa HCO₃^?/CO₃^2? = 10.30)

c) NaHCO₃ (0.20 M) (use pKa values from part "b" above)

a

b

c

2.

Calculate the approximate pH of the following aqueous solutions at 25C. (Note: consider one decimal place for the pH answer.)

a) NH₄Cl (0.20 M) + CH₃COONa (0.40M) , (pK_a NH₄⁺/NH₃ = 9.24, pK_aCH₃COOH/CH₃COO^? = 4.76)

b) NH₃ (0.040 M)+ NaOH (0.040M)

a

b

3.

Calculate the approximate pH of the following aqueous solutions at 25C. (Note: consider one decimal place for the pH answer.)

a) H₂S (0.10 M) + H₃PO₄ (0.20M) , (pK_aH₂S/HS^? = 7.02, pK_a HS^?/S^2? = 13.99, and H₃PO₄ pK_a values are given in part "a" of question 1)

b) H₃PO₄ (0.20M) + NaH₂PO₄ (0.10 M) (use pKa values given in question 1)

c) NH₄H₂PO₄ (0.1 0M) (pKa values are given in question 1 and 2 )

a

b

c

4.

A chemist prepares a solution that contains a formal concentration of 0.5 M phthalic acid and that has a pH of 2.88. What is the approximate fraction of phthalic acid in this solution at 25 C? Use 1 sig figs for the answer

Phthalic Acid: HOOCC₆H₄COOH/HOOCC₆H₄COO^- pKa1 = 2.95, HOOCC₆H₄COO^-/^-OOCC₆H₄COO^- pKa₂ = 5.41

Bonus: For more practice find the fractions of other species (forms) at this pH and decide which one is the predominant form. Plot an approximate curve of alpha vs pH for each species. You can hand them in for extra credit.

5.

How many grams of Ba(IO₃)₂ (FW=487 g/mol) can be dissolved in one liter of the following solutions at 25C? (Ba(IO₃)₂K_sp=1.57x10^?9) Use three significant figures for the answer. You do not need to consider the effect of electrolyte.

a) In water.

b) In a solution of Ba(NO₃)₂ (0.0700M)

a

b

6.

If a 0.100 M solution of NaOH is added to a solution containing 0.200 M Ni²⁺, 0.200 M Ce³⁺, and 0.200 M Cu²⁺, which metal hydroxides will precipitate first and last, respectively? K_sp for Ni(OH)₂ is 6.0x10^?16, for Ce (OH)₃ is 6.0 x 10^?22, and for Cu(OH)₂ is 4.8x10^?20.

A) Ni(OH)₂ first, Cu(OH)₂ last
B) Cu(OH)₂ first, Ni(OH)₂ last
C) Ni(OH)₂ first, Ce(OH)₃ last
D) Cu(OH)₂ first, Ce(OH)₃ last
E) Ce(OH)₃ first, Cu(OH)₂ last

Give detailed explanation

Answer 1

Part 1

a) H3PO4 ----> H2PO4- + H+

       0.2M                 0         0

0.2-x                 x          x

pKa1= -log Ka1 ---> Ka1= 7.08 x 10^-3 --> Ka1=[H+][H2PO4-]/[H3PO4]= x²/ 0.2-x ? x= 0.0343M

H2PO4- ----> HPO4-2 + H+

0.0343M        0             0.0343M

0.0343-x         x             0.0343 + x

pKa2= -log Ka2 --> Ka2= 7.08 x 10^-8= [H+][HPO4-2]/[H2PO4-]=(0.0343+x) x/0.0343-x

x=7.08 x 10^-8 M

HPO4-2 ----------------> H+        + PO4-3

7.08 x10^-8 M 0.0343      0

7.08 x10^-8 -x 0.0343+x   x

pKa3= -log Ka3 --> Ka3= 4.17 x 10^-13= [H+][PO4-3]/[HPO4-2]=(0.0343+x)x/7.08x10^-8-x

x=8.6 x10^-19M

pH= -log[H+]= -log 0.0343= 1.45

b) Na2CO3 is a salt made from NaOH and H2CO3. Na+ is the conjugated acid of NaOH a strong base so it does not affect the pH of the solution. CO3-2 is the conjugated base of H2CO3 a weak acid, so it will affect the pH, the final solution will be basic.

CO3-2 + H2O----> HCO3- + OH-

0.2M                          0        0

0.2-x                          x          x

We have pKa, so we can calculate Ka for HCO3-/CO3-2 but we need Kb. First calculate Ka:

pKa= -logKa ----> Ka= 5.01 x 10^-11 ---> Ka x Kb=Kw ---> Kb= 1 x 10^-14/ 5.01 x 10^-11= 2 x 10^-4

Kb=[HCO3-][OH-]/[CO3-2]= x²/0.2-x -----> x= 6.23x10^-3M

HCO3- + H2O ----> H2CO3 + OH-

6.23x10^-3                   0         6.23x10^-3

6.23x10^-3-x                x          6.23x10^-3+x

In this case we have to do the same than the first part:

pKa= -logKa ----> Ka= 4.47 x 10^-7 --->Ka x Kb=Kw ---> Kb= 1 x 10^-14/4.47 x 10^-7= 2.24 x 10^-8

Kb=[H2CO3][OH-]/[HCO3-]= x (6.23x10^-3+x)/6.23x10^-3-x ---->x=2.24 x 10^-8M

pOH= -log[OH-]= -log (6.23x10^-3+2.24 x10^-8) ? -log 6.23x10^-3 = 2.2

pH +pOH= 14----> pH= 11.8

c) Na+ is the conjugated acid of NaOH so, again, it does not affect pH.

HCO3- + H2O ----> H2CO3 + OH-

0.2                              0         0

0.2-x                            x        x

pKa= -logKa ----> Ka= 4.47 x 10^-7 --->Ka x Kb=Kw ---> Kb= 1 x 10^-14/4.47 x 10^-7= 2.24 x 10^-8

Kb=[H2CO3][OH-]/[HCO3-]= x²/0.2-x ---> x= 6.7 x10^-5M

pOH= -log[OH-] = -log 6.7 x 10-5 = 4.2--->pH+pOH=14---> pH= 9.8

Part 2

a) Na+ and Cl- does not affect pH because they are the conjugated acid and base of a strong base and strong acid respectively.Then we have NH4+, the conjugated acid of a weak base, so it will affect pH; CH3COO- is the conjugated base of a weak acid so it will affect pH too.

NH4+ + CH3COO- -----> NH4CH3COO + H2O

I will assume that the volume of the solution is 1L.

mol NH4+= 0.2M x 1L= 0.2 mol

mol CH3COO-= 0.4M x 1L= 0.4 mol

1 mol NH4+ -------- 1 mol CH3COO-

0.2 mol NH4+ -----x= 0.2 mol CH3COO- ----> we have 0.4 mol of CH3COO- so it is in excess.

This means that what is really happening is:

NH4+ + CH3COO- -----> NH4CH3COO + H2O + CH3COO-

The excess of CH3COO- will react with water:

CH3COO- + H2O ---> CH3COOH + OH-

0.2M                                 0              0

0.2-x                                   x             x

pKa= -logKa ----> Ka= 1.74 x 10^-5 --->Ka x Kb=Kw ---> Kb= 5.75 x 10^-10

Kb=[CH3COOH][OH-]/[CH3COO-] = x²/0.2-x----> x= 1.07 x10^-5M

pOH= -log[OH-]= 5.0 ----> pH + pOH= 14 ---> pH= 9.0

b) NH3 + H2O< ----> NH4+ + OH-

If we add NaOH the equilibrium will shift to the left because we are adding OH-.

First find the equilibrium concentrations of all species before the addition of NaOH:

NH3 + H2O< ----> NH4+ + OH-

0.04                        0         0

0.04-x                     x           x

pKa= -logKa= 9.24 ----> Ka= 5.75 x 10^-10 --->Ka x Kb=Kw ---> Kb= 1.74 x 10^-5

Kb=[NH4+][OH-]/[NH3]= x²/0.04-x ---> x= 8.26 x10^-4M

[NH3]equilibrium= 0.04-8.26 x10-4= 0.0392M

[OH-]equilibrium= 8.26 x10^-4M

NH3 + H2O< ----> NH4+        +           OH-

0.0392 8.26 x10^-4     8.26 x10^-4

0.0392 +x         8.26 x10^-4-x     (8.26 x10^-4+0.04) -x

Kb= ((8.26 x10^-4+0.04) -x)(8.26 x10^-4-x)/0.0392 +x ---> x= 8.25x10^-4M

[OH-]?0.04M---->pOH=-log[OH-]= 1.4 --->pH=14-pOH=12.6

Part 3

a)H2S and H3PO4 are weak acids, we need to do an ICE table to find the [H+] of both acids.

Let