CHEM 212:CHEMISTSRY V 2
CHEM 212:CHEMISTSRY V 2
8th Edition
ISBN: 9781260304503
Author: SILBERBERG
Publisher: MCGRAW-HILL CUSTOM PUBLISHING
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Chapter 18, Problem 18.149P

(a)

Interpretation Introduction

Interpretation:

Carbon dioxide when dissolved in water undergoes multistep equilibrium process.  The reactions are,

      CO2(g) + H2O(l)H2CO3(aq)H2CO3(aq) + H2O(l)HCO3-(aq) + H3O+(aq)

Each step has to be classified as a Lewis or a Bronsted-Lowry reaction.

Concept Introduction:

Bronsted-Lowry reaction:

Any species that has the capability of accepting a proton, which requires a lone pair of electrons to bond to H+ is said to be Bronsted-Lowry base.

Any species that has the capability of donating a proton (H+) is said to be Bronsted-Lowry acid.

According to Bronsted-Lowry theory, an acid-base reaction in which a proton is transferred from an acid to base is said to be Bronsted-Lowry reaction.

Example:

  NH3(g)+ HCl(g)  NH4Cl(s)

In the above reaction, HCl acts as Bronsted-Lowry acid and donates proton which is accepted by NH3 using its lone pair.  So, NH3 is Bronsted-Lowry base.

Lewis reaction:

In an acid-base reaction, Lewis base donates electrons to the acid and Lewis acid accepts the electron pair to form a covalent bond between Lewis acid and Lewis base.

Example:

  A + B  A-B

In the above example, A is an acid and B is a base.  B donates electrons to A and forms an adduct.

(a)

Expert Solution
Check Mark

Explanation of Solution

The given reactions are

      CO2(g) + H2O(l)H2CO3(aq)H2CO3(aq) + H2O(l)HCO3-(aq) + H3O+(aq)

Lewis reaction:

  CO2(g) + H2O(l)H2CO3(aq)

In the formation of carbonic acid, CO2 is the Lewis acid and H2O is Lewis base.  H2O donates electrons to CO2 to form H2CO3.

Bronsted-Lowry reaction:

  H2CO3(aq) + H2O(l)HCO3-(aq) + H3O+(aq)

The above reaction is Bronsted-Lowry reaction and also Lewis reaction.

(b)

Interpretation Introduction

Interpretation:

The pH value of non-polluted rainwater in equilibrium with clean air has to be calculated.  (PCO2 in clean air = 4×10-4 atm; Henry’s law constant for CO2 at 25°C is 0.033 mol/L.atm and Koverall = 4.5×10-7).

(b)

Expert Solution
Check Mark

Explanation of Solution

The reaction can be given as

  CO2(g)+ 2H2O(l)HCO3-(aq)+ H3O+(aq)

The molarity of CO2 can be calculated as

  Molarity of CO2 = kHPCO2Molarity of CO2 = (0.033 mol/L.atm)(4×10-4atm)    = 1.320×10-5 M CO2

Given: Koverall = 4.5×10-7,

  Koverall = [H3O+][HCO3-][CO2]4.5×10-7 = [x][x][1.320×10-5-x]

Assume x is small when compared to 1.320×10-5.

  4.5×10-7 = [x][x][1.320×10-5]x = 2.4372×10-6

The value of x calculated has to be compared to 1.320×10-5.

  2.4372×10-61.320×10-5 (100) = 18% error

As the calculated error is more than 5%, the assumption of ‘x’ value smaller than the value of 1.320×10-5 is not acceptable.

Quadratic equation is necessary to use to calculate x.

  x2 = (4.5×10-7)(1.320×10-5-x)x2 = 5.940×10-12- 4.5×10-7xx2 + 4.5×10-7x - 5.940×10-12 = 0

Consider a = 1, b = 4.5×10-7, c = -5.940×10-12.

  x = -b ± b2- 4ac2ax = -4.5×10-7 ± (4.5×10-7)2- 4(1)(-5.940×10-12)2(1)x = 2.222575 × 10-6 M H3O+

The pH can be calculated as

  pH = -log[H3O+]pH = -log (2.222575 × 10-6pH = 5.6531

The pH of non-polluted rainwater in equilibrium with clean air is 5.6.

(c)

Interpretation Introduction

Interpretation:

The [CO32-] in rain water has to be calculated.  (Ka of  HCO3- = 4.7 × 10-11).

(c)

Expert Solution
Check Mark

Explanation of Solution

The reaction can be given as

  HCO3-(aq)+ H2O(l)H3O+(aq)+ CO32-(aq)

Given, Ka of  HCO3- = 4.7 × 10-11.

  Ka = [H3O+][CO32-][HCO3-]

Using the x value calculated in sub-part (b), x = 2.222575 × 10-6 M H3O+,

  4.7 × 10-11 = [2.222575 × 10-6+ x][x][2.222575 × 10-6- x]assume x is small compared to 2×10-6,4.7 × 10-11 = [2.222575 × 10-6][x][2.222575 × 10-6][CO32-] = x = 4.7×10-11 = 5×10-11 M CO32-

The value of x is compared to 2×10-6.

  4.7×10-112.222575×10-6 (100) = 0.0021% error

As the calculated error is less than 5% ; the assumption made is valid.

The [CO32-] in rain water is calculated as 5×10-11.

(d)

Interpretation Introduction

Interpretation:

If the partial pressure of CO2 in clean air doubles in the next few decades, the pH of rainwater has to be calculated.

(d)

Expert Solution
Check Mark

Explanation of Solution

Given, PCO2 in clean air = 4×10-4 atm; Henry’s law constant for CO2 at 25°C is 0.033 mol/L.atm and Koverall = 4.5×10-7.

The value of PCO2 is doubled.  The molarity of CO2 can be calculated as

  Molarity of CO2 = 2kHPCO2Molarity of CO2 = 2(0.033 mol/L.atm)(4×10-4atm)    = 2.640 ×10-5 M CO2

Given: Koverall = 4.5×10-7,

  Koverall = [H3O+][HCO3-][CO2]4.5×10-7 = [x][x][2.640×10-5-x]

Assume x is small when compared to 2.640×10-5.

  4.5×10-7 = [x][x][2.640×10-5]x = 3.44674×10-6

The value of x calculated has to be compared to 2.640×10-5.

  3.44674×10-62.640×10-5 (100) = 13% error

As the calculated error is more than 5%, the assumption ‘x’ smaller than the value of 2.640×10-5 is not acceptable.

Quadratic equation is necessary to use to calculate x.

  x2 = (4.5×10-7)(2.640×10-5-x)x2 = 1.1880×10-11- 4.5×10-7xx2 + 4.5×10-7x - 1.1880×10-11 = 0

Consider: a = 1, b = 4.5×10-7, c = -1.1880×10-11.

  x = -b ± b2- 4ac2ax = -4.5×10-7 ± (4.5×10-7)2- 4(1)(-1.1880×10-11)2(1)x = 3.229 × 10-6 M H3O+

The pH can be calculated as

  pH = -log[H3O+]pH = -log (3.229 × 10-6pH = 5.4909 = 5.5

The value of pH rainwater when partial pressure of CO2 doubled is 5.5.

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Chapter 18 Solutions

CHEM 212:CHEMISTSRY V 2

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