Solutions to: Acid-Base and Solubility Homework Problem Set

[Pages:30]Solutions to: Acid-Base and Solubility Homework Problem Set

S.E. Van Bramer 1/3/17

1.Calculate the H3O1+ concentration, OH1- concentration, pH, and pOH of the following solutions. First solve assuming that Kw is insignificant. Repeat the calculations and include Kw (you will need to use the quadratic equation to solve these). When is Kw significant?

M := moleliter- 1

Kw := 1.010- 14M2

1-a.1.0 x 10-9 M HNO3 Cacid := 1.010- 9M

Assuming Kw is insignificant: CH3O := Cacid

pH := -logCH3OM- 1

pOH := 14 - pH

CH3O

=

1

?

- 10

9M

pH = 9

Does this make sense?

pOH = 5

Assuming Kw is significant (Included the initial [H3O1+] in the calculations.)

CH3O := Kw + Cacid

CH3O

=

1.01

?

- 10

7M

pH := -logCH3OM- 1

pH = 6.996

pOH := 14 - pH

pOH = 7.004

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

1-b.1.0 x 10-6 M HNO3 Cacid := 1.010- 6M

Assuming Kw is insignificant: CH3O := Cacid

pH := -logCH3OM- 1

pOH := 14 - pH

CH3O

=

1

?

- 10

6M

pH = 6

pOH = 8

Assuming Kw is significant (Included the initial [H3O1+] in the calculations.)

CH3O := Kw + Cacid

CH3O

=

1.1 ?

- 10

6M

pH := -logCH3OM- 1

pH = 5.959

pOH := 14 - pH

pOH = 8.041

1-c.10.0 M HNO3 Cacid := 10M

Assuming Kw is insignificant: CH3O := Cacid

pH := -logCH3OM- 1

pOH := 14 - pH

CH3O = 10 M pH = -1 pOH = 15

Notice that pH can be negative. A t concentrations this high, however, the assumption that all HNO3 dissociates is probably not accurate.

Assuming Kw is significant (Included the initial [H3O1+] in the calculations.)

CH3O := Kw + Cacid

pH := -logCH3OM- 1

pOH := 14 - pH

CH3O = 10 M pH = -1 pOH = 15

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

2. Calculate the H3O1+ concentration, OH1- concentration, pH, pOH, nitrate ion concentration and sodium ion concentration for the following experimental steps in a titration. Graph your results.

2-a.6 M HNO3. This is just a strong acid. Cacid := 6M

CH3O := Cacid

pH := -logCH3OM- 1

CH3O = 6 M pH = -0.778

pOH := 14 - pH

pOH = 14.778

2-b.1.00 mL of the nitric ac id is diluted with deionized water to a volume of 100.0 ml.

The concentration of the acid is : Cstock := 6M

Vstock := 1.00mL

Vacid := 100mL

Cacid :=

C stock Vstock Vacid

Cacid = 0.06 M

The concentrations: CH3O := Cacid

pH := -logCH3OM- 1

pOH := 14 - pH

CH3O = 0.06 M pH = 1.222 pOH = 12.778

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

2-c.5.2468 g of NaOH is dissolved in deionized water to a volume of 1.000 liter. The concentration of sodium hydroxide is:

massbase := 5.2468gm MWbase := (22.990 + 15.999 + 1.0079)gmmole- 1

molebase :=

massbase MWbase

volumebase := 1000mL

Cbase :=

molebase volumebase

Cbase = 0.1312 M

The concentrations:

COH := Cbase

pOH

:=

-logCbase

- M

1

pH := 14 - pOH

COH = 0.131 M pOH = 0.882 pH = 13.118

2-d.1.00 mL of the sodium hydroxide solution is added to the nitric ac id solution.

Vacid := 100mL

Vbase := 1.00mL

Vtotal := Vacid + Vbase

Vtotal = 101 mL

Calculate the pH by assuming that all the added OH-, reacts with H3O already present.

moleH3O := CacidVacid - CbaseVbase

moleH3O

=

5.869

?

- 10

3 mole

hwkc16_a.xmcd

CH3O :=

moleH3O Vtotal

COH :=

Kw CH3O

pH

:=

-1

log

CH3O

- M

1

CH3O = 0.058 M

COH

=

1.721

?

- 10

13 M

pH = 1.236

pOH := 14 - pH

pOH = 12.764

CNO3 :=

CacidVacid Vtotal

CNa :=

CbaseVbase Vtotal

CNO3 = 0.059 M

CNa

=

1.299

?

- 10

3M

S.E. Van Bramer

1/3/2017

2-e.20.00 mL of the s odium hydroxide solution is added to the nitric acid s olution.

Vacid := 100mL Vtotal := Vacid + Vbase Vtotal = 120 mL

Vbase := 20mL

Calculate the pH by assuming that all the added OH-, reacts with H3O already present.

moleH3O := CacidVacid - CbaseVbase

moleH3O

=

3.376

?

- 10

3 mole

CH3O :=

moleH3O Vtotal

COH :=

Kw CH3O

pH

:=

-1

log

CH3O

- M

1

pOH := 14 - pH

CNO3 :=

CacidVacid Vtotal

CH3O = 0.028 M

COH

=

3.554

?

- 10

13 M

pH = 1.551 pOH = 12.449 CNO3 = 0.05 M

CNa :=

CbaseVbase Vtotal

CNa = 0.022 M

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

2-f.Enough sodium hydroxide solution is added to reach the equivilence point.

At the equivalence point enough base is added so that the moles of hydroxide and the moles of nitric acid are equivilent.

moleacid := CacidVacid molebase := moleacid

moleacid

=

6

?

- 10

3 mole

molebase

=

6

?

- 10

3 mole

Vbase :=

molebase Cbase

Vbase = 45.739 mL

Vtotal := Vacid + Vbase

Vtotal = 145.739 mL

Calculate the pH by assuming that all the added OH-, reacts with H3O already present.

moleH3O := CacidVacid - CbaseVbase

moleH3O = 0 mole

Since the acid and base do not contribute to the pH, it is the same as pure water: For pure water, the Kw equlibrium determines the pH

Kw = CH3OCOH

CH3O = X

COH = X

X := Kw

CH3O := X

pH := -logCH3OM- 1

pOH := 14 - pH

CH3O

=

1

?

- 10

7M

pH = 7 pOH = 7

CNO3 :=

CacidVacid Vtotal

CNO3 = 0.041 M

CNa :=

CbaseVbase Vtotal

CNa = 0.041 M

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

2-g. .00 mL of the sodium hydroxide solution is added to the nitric acid solution.

Vacid := 100mL Vtotal := Vacid + Vbase Vtotal = 150 mL

Vbase := 50mL

Calculate the pH by assuming that all the added OH-, reacts with H3O already present. moleH3O := CacidVacid - CbaseVbase

moleH3O

=

-5.59

?

- 10

4 mole

This result does not make any sense because nitric acid is now the limiting

reagent. So OH1- is in excess and determines the pH

moleOH := CbaseVbase - CacidVacid

COH :=

moleOH Vtotal

pOH

:=

-logCOH

- M

1

pH := 14.0 - pOH

CNO3 :=

CacidVacid Vtotal

CNa :=

CbaseVbase Vtotal

moleOH

=

5.59

?

- 10

4 mole

COH

=

3.727

?

- 10

3M

pOH = 2.429 pH = 11.571 CNO3 = 0.04 M

CNa = 0.044 M

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

3.What is the pH and pOH of a buffer prepared by adding 1.2435 g of sodium acetate to 100 mL of 0.124

M acetic acid. Calculate using the Henderson-Hasselbalch equation (as shown in your textbook) and using

the quadratic equation (as shown in class).

Ka

:=

1.8

- 10

5M

The concentration of the acid in the buffer is: Cacid := 0.124M

The concentration of the salt is (Fill in appropriate values and delete unnecessary sections.): masssalt := 1.2435gm MWsalt := [[(22.990) + (212.001) + (31.0079)] + (215.999)]gmmole- 1

molesalt :=

masssalt MWsalt

Vsalt := 100mL

Csalt :=

molesalt Vsalt

molesalt = 0.015 mole Csalt = 0.152 M

The equlibrium solution for Initial Concentraion

HA + Cacid

H2O

H3O+ + ACsalt

Equlibrium Concentration

Cacid - X

X

Csalt + X

Equlibrium expression:

( ) Csalt + X X

Ka = Cacid - X

Assuming X is smaller than [acid] and [salt], this reduces to:

( ) Csalt (X) ( ) Ka = Cacid

Solving for X:

X := KaCacid Csalt

X

=

1.4721

?

- 10

5

M

hwkc16_a.xmcd

S.E. Van Bramer

1/3/2017

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download