Chemistry CHM 1310D Test 2November 24, 2011

Darrin Richeson

There were 3 versions to the exam and they appear sequentially on the following pages.

1. A 2.85 g sample of a base symbolized by B, with a molar mass of 45.09 g/mol, was dissolved in water to produce 100.0 ml of solution with a pH = 11.53. Write the equation for the reaction of B in water and calculate the ionization constant (Kb) for this base. What is the equilibrium constant for the following reaction?

BH+ +H2O → B + H3O+

Answer:

B + H2O → BH+ + OH-

2.85g/45.09g/mol = 6.321 x 10-2; 0.632M

pH = 11.53 pOH = 14-11.53 = -log [OH-] = 2.47 [OH-] = 3.388 x 10-3

B + H2O → BH+ + OH-

I 0.632

C -3.388 x 10-3 3.388 x 10-3 3.388 x 10-3

E 0.6286 3.388 x 10- 3.388 x 10-3

(3.388 x 10-3)2/(0.6286) = Kb = 1.826 x 10-5

The reaction corresponds to Ka of the conjugate acid of base B. Calculated from the relationship Kw = Ka x Kb for a conjugate pair.

Thus Ka = 10-14/1.826 x 10-5 = 5.47 x 10-10

1

2. You are given a 0.500L of a 0.660M solution of HOBr (Ka= 2.00x10-9)(a) What is the pH of this solution?

Answer:

HOBr → H+ OBr-

0.660 - -0.660-x x x

x2/0.660-x = 2.00x 10-9

Since 0.66>>2x10-9 we will use the “assumption” and 0.66-x = 0.66x2 = 2.00x10-9 x 0.660 = 1.32x10-9

x = 3.63 x 10-5

This is the [H+] so pH = 4.44

(b) You now add 18.51g of KOH to this solution. What is the pH of this solution?

Answer:Moles of HOBr = 0.500L x 0.660M = 0.330 moles HOBr

18.5 g x (1 mole/56.11g) = 0.3297 = 0.330 moles Added 0.330 moles of KOH

The reaction is 1:1KOH + HOBr → K+ + H2O + OBr-

This solution now contains 0.330 mole/0.5L or 0.660M OBr-

This species is a baseOBr- + H2O → HOBr +OH- and Kb = Kw/Ka = 10-14/2.00x10-9 = 0.500 x 10-5

OBr- + H2O HOBr + OH- 0.6600.660-x x x

x2/(0.660-x) = 5.00x10-6

Since 0.660>>5x10-6 we will use the “assumption” and 0.66-x = 0.66x2 = 0.660 x 5x10-6 x = [OH-] = 1.82 x 10-6 pOH = 2.74 and pH = 11.26

2

3. You are given a weak acid with the formula HAn (122.1g/mol). When you make a solution of 0.500 g of HAn and 0.646 g Ca(An)2 in 0.500 L of water the pH is 5.92.(a) Calculate Ka for the acid.(b) If 2.0 mL of 1.0 M HCl are added to this solution what will be the pH?(c) How many grams of NaOH should be added to this solution to get a pH of 6.25.

Answer:(a) This is a mixture of a weak acid and its conjugate base

Ka = [An-][H+]/[HAn] [HAn]= 0.500 x 1mol/122.1g in 0.500L = 4.095 x 10-3 moles/0.500L [An-] = 0.646 x 1mol/282.1gx 2moles [An-]/mole of Ca salt in 0,500L

= 4.58 x 10-3 moles/0.500L[H+] = 10-5.92= 1.202 x 10-6

moles/L

And Ka = 1.34 x 10-6 = 4.58x 10-3x 1.202x10-6 /4.095x10-3

Could use the equation pH = pKa +log [A]/[HA] and will get the same number.

(b) have added 2.0 x10-3 L x 1mole/L = 2.0 x 10-3 moles HCl. This reacts with An- to yield HAn. Convert 2.0 x 10-3 moles of An- to HAn. New concentrations are

(4.095x 10-3 + 2.0 x 10-3)moles HAn/0.502L = 6.095x 10-3moles/0.502L

(4.58x 10-3 – 2.0 x 10-3)moles An-/0.502L = 2.58 x 10-3moles/0.502L

[H+] = Ka [HAn]/[An-] = 3.166 x 10-6

pH = 5.50

(c) For a pH =6.25, [H+] = 5.62 x 10-7M

the ratio[HAn]/[An-] = [H+]/Ka = 0.4197 = (4.095 x 10-3-x)/4.58 x 10-3 + x)

need to add enough OH- to adjust this ratio. x moles of OH- makes x moles of An- from x moles HAn.

Solve for x:

1.922 x 10-3 + 0.4197 x = 4.095x 10-3 –x

1.4197 x = 2.173x 10-3

x = 1.531 x 10-3 moles

need 1.531 x 10-3 moles x 40g NaOH/mole = 0.0612 g

3

4. (a) What is the minimum value of n for the following l values?l = 2 minimum n = 3

l = 3 minimum n = 4

(b) Circle (neatly!) the best single answer to the following question.

An atomic wavefunction represents:

i. the exact location of the electron

ii. the repulsion of all the electrons among themselves

iii. the region of high probability for an electron around the nucleus of an atom

iv. the region of high electron density for a covalent bond

v. an orbit that an electron follows around the nucleus of an atom

(c) How many orbitals are available with the following sets of quantum numbers?n = 4, l = 2, ml = 1 There is only one orbital (2 electrons)

n = 3, l = 3, ml = -1, ms= -1/2 zero orbitals, for n = 3 l cannot equal 3

n = 4, l = 3 ml =-3, -2, -1, 0, 1, 2, 3. There are 7 orbitals (they can hold 14 electrons)

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5. For the reaction A → C, you are told that increasing the [A] by a factor of three leads to an increase in the reaction rate by a factor of 3. You are also given the following data.

[A] M time (min) 0.400 00.250 15.0

Propose a form for the rate law?What is the value of the rate constant, k? What is the half-life of the reaction? How long will it take for ¾ of the original concentration of A to be consumed?

Answer:

The tripling of the reaction rate with triple concentration indicate that the rate law is first order in A

Rate of reaction = k[A]

Using the integrated rate law provided:ln[A]t – ln[A]0 = -kt = -k(15min) = ln 0.250 – ln 0.400 = -0.700

k = 0.0313 min-1

The half-life, t1/2 is the time when half of the original [A] has been consumed. Either use ln[A]t – ln[A]0 = -kt = -0.0313 min -1(t1/2) = ln 0.200 – ln 0.400 = -0.693 (t1/2) = -0.693/-0.0313 min -1 = 22.1 minor t1/2 = ln2/k = 0.693/0.0313(these are the same!)

Can either use the integrated rate law or realize that ¾ is 2 x t1/2

ln[A]t – ln[A]0 = -kt = -0.0313 min -1(t) = ln 0.1 – ln 0.400 = -1.386

44.3 min.

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5. A 2.65 g sample of a base symbolized by B, with a molar mass of 45.09 g/mol, was dissolved in water to produce 100.0 ml of solution with a pH = 11.53. Write the equation for the reaction of B in water and calculate the ionization constant (Kb) for this base. What is the equilibrium constant for the following reaction?

BH+ +H2O → B + H3O+

Answer:

B + H2O → BH+ + OH-

2.65g/45.09g/mol = 5.877 x 10-2; 0.5877M

pH = 11.53 pOH = 14-11.53 = -log [OH-] = 2.47 [OH-] = 3.388 x 10-3

B + H2O → BH+ + OH-

I 0.5877

C -3.388 x 10-3 3.388 x 10-3 3.388 x 10-3

E 0.5843 3.388 x 10- 3.388 x 10-3

(3.388 x 10-3)2/(0.5843) = Kb = 1.97 x 10-5

The reaction corresponds to Ka of the conjugate acid of base B. Calculated from the relationship Kw = Ka x Kb for a conjugate pair.

Thus Ka = 10-14/1.97 x 10-5 =5.08x 10-10

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6. You are given a 0.500L of a 0.660M solution of HOBr (Ka= 1.90x10-9)(c) What is the pH of this solution?

Answer:

HOBr → H+ OBr-

0.660 - -0.660-x x x

x2/0.660-x = 1.90x 10-9

Since 0.66>>1.90x10-9 we will use the “assumption” and 0.66-x = 0.66x2 = 1.90x10-9 x 0.660 = 1.24x10-9

x = 3.54 x 10-5

This is the [H+] so pH = 4.45

(d) You now add 18.51g of KOH to this solution. What is the pH of this solution?

Answer:Moles of HOBr = 0.500L x 0.660M = 0.330 moles HOBr

18.5 g x (1 mole/56.11g) = 0.3297 = 0.330 moles Added 0.330 moles of KOH

The reaction is 1:1KOH + HOBr → K+ + H2O + OBr-

This solution now contains 0.330 mole/0.5L or 0.660M OBr-

This species is a baseOBr- + H2O → HOBr +OH- and Kb = Kw/Ka = 10-14/1.90x10-9 = 5.26 x 10-6

OBr- + H2O HOBr + OH- 0.6600.660-x x x

x2/(0.660-x) = 5.26x10-6

Since 0.660>>5.26x10-6 we will use the “assumption” and 0.66-x = 0.66x2 = 0.660 x 5.26x10-6 x = [OH-] = 1.86x 10-3 pOH = 2.72 and pH = 11.27

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7. You are given a weak acid with the formula HAn (122.1g/mol). When you make a solution of 0.500 g of HAn and 0.646 g Ca(An)2 in 0.500 L of water the pH is 5.92.(a) Calculate Ka for the acid.(b) If 2.0 mL of 1.0 M HCl are added to this solution what will be the pH?(c) How many grams of NaOH should be added to this solution to get a pH of 6.25.

Answer:(a) This is a mixture of a weak acid and its conjugate base

Ka = [An-][H+]/[HAn] [HAn]= 0.500 x 1mol/122.1g in 0.500L = 4.095 x 10-3 moles/0.500L [An-] = 0.646 x 1mol/282.1gx 2moles [An-]/mole of Ca salt in 0,500L

= 4.58 x 10-3 moles/0.500L[H+] = 10-5.92= 1.202 x 10-6

moles/L

And Ka = 1.34 x 10-6 = 4.58x 10-3x 1.202x10-6 /4.095x10-3

Could use the equation pH = pKa +log [A]/[HA] and will get the same number.

(b) have added 2.0 x10-3 L x 1mole/L = 2.0 x 10-3 moles HCl. This reacts with An- to yield HAn. Convert 2.0 x 10-3 moles of An- to HAn. New concentrations are

(4.095x 10-3 + 2.0 x 10-3)moles HAn/0.502L = 6.095x 10-3moles/0.502L

(4.58x 10-3 – 2.0 x 10-3)moles An-/0.502L = 2.58 x 10-3moles/0.502L

[H+] = Ka [HAn]/[An-] = 3.166 x 10-6

pH = 5.50

(c) For a pH =6.25, [H+] = 5.62 x 10-7M

the ratio[HAn]/[An-] = [H+]/Ka = 0.4197 = (4.095 x 10-3-x)/4.58 x 10-3 + x)

need to add enough OH- to adjust this ratio. x moles of OH- makes x moles of An- from x moles HAn.

Solve for x:

1.922 x 10-3 + 0.4197 x = 4.095x 10-3 –x

1.4197 x = 2.173x 10-3

x = 1.531 x 10-3 moles

need 1.531 x 10-3 moles x 40g NaOH/mole = 0.0612 g

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8. (a) What is the minimum value of n for the following l values?l = 1 minimum n = 2

l = 2 minimum n = 3

(b) Circle (neatly!) the best single answer to the following question.

An atomic wavefunction represents:

i. the exact location of the electron

ii. the repulsion of all the electrons among themselves

iii. the region of high electron density for a covalent bond

iv. an orbit that an electron follows around the nucleus of an atom

v. the region of high probability for an electron around the nucleus of an atom

(c) How many orbitals are available with the following sets of quantum numbers?n = 4, l = 2, ml = 1 There is only one orbital (2 electrons)

n = 3, l = 3, ml = -1, ms= -1/2 zero orbitals, for n = 3 l cannot equal 3

n = 4, l = 3 ml =-3, -2, -1, 0, 1, 2, 3. There are 7 orbitals (they can hold 14 electrons)

9

9. For the reaction A → C, you are told that increasing the [A] by a factor of three leads to an increase in the reaction rate by a factor of 3. You are also given the following data.

[A] M time (min) 0.400 00.250 9.00

Propose a form for the rate law?What is the value of the rate constant, k? What is the half-life of the reaction? How long will it take for ¾ of the original concentration of A to be consumed?

Answer:

The tripling of the reaction rate with triple concentration indicate that the rate law is first order in A

Rate of reaction = k[A]

Using the integrated rate law provided:ln[A]t – ln[A]0 = -kt = -k(9min) = ln 0.250 – ln 0.400 = -0.4700

k = 0.0522 min-1

The half-life, t1/2 is the time when half of the original [A] has been consumed. Either use ln[A]t – ln[A]0 = -kt = -0.0522 min -1(t1/2) = ln 0.200 – ln 0.400 = -0.693 (t1/2) = -0.693/-0.0313 min -1 = 13.3 minor t1/2 = ln2/k = 0.693/0.0522(these are the same!)

Can either use the integrated rate law or realize that ¾ is 2 x t1/2

ln[A]t – ln[A]0 = -kt = -0.0522 min -1(t) = ln 0.1 – ln 0.400 = -1.386

26.6 min.

10

10. A 2.85 g sample of a base symbolized by B, with a molar mass of 45.09 g/mol, was dissolved in water to produce 100.0 ml of solution with a pH = 11.33. Write the equation for the reaction of B in water and calculate the ionization constant (Kb) for this base. What is the equilibrium constant for the following reaction?

BH+ +H2O → B + H3O+

Answer:

B + H2O → BH+ + OH-

2.85g/45.09g/mol = 6.321 x 10-2; 0.632M

pH = 11.33 pOH = 14-11.33 = -log [OH-] = 2.67 [OH-] = 2.138 x 10-3

B + H2O → BH+ + OH-

I 0.632

C -2.138 x 10-3 2.138 x 10-3 2.138 x 10-3

E 0.6299 2.138 x 10- 2.138 x 10-3

(2.138 x 10-3)2/(0.6299) = Kb = 7.257 x 10-6

The reaction corresponds to Ka of the conjugate acid of base B. Calculated from the relationship Kw = Ka x Kb for a conjugate pair.

Thus Ka = 10-14/7.257 x 10-6 = 1.378 x 10-9

11

11. You are given a 0.500L of a 0.660M solution of HOBr (Ka= 2.10x10-9)(e) What is the pH of this solution?

Answer:

HOBr → H+ OBr-

0.660 - -0.660-x x x

x2/0.660-x = 2.10x 10-9

Since 0.66>>2.1x10-9 we will use the “assumption” and 0.66-x = 0.66x2 = 2.10x10-9 x 0.660 = 1.39x10-9

x = 3.72 x 10-5

This is the [H+] so pH = 4.43

(f) You now add 18.51g of KOH to this solution. What is the pH of this solution?

Answer:Moles of HOBr = 0.500L x 0.660M = 0.330 moles HOBr

18.5 g x (1 mole/56.11g) = 0.3297 = 0.330 moles Added 0.330 moles of KOH

The reaction is 1:1KOH + HOBr → K+ + H2O + OBr-

This solution now contains 0.330 mole/0.5L or 0.660M OBr-

This species is a baseOBr- + H2O → HOBr +OH- and Kb = Kw/Ka = 10-14/2.10x10-9 = 0.476 x 10-5

OBr- + H2O HOBr + OH- 0.6600.660-x x x

x2/(0.660-x) = 4.76x10-6

Since 0.660>>4.76x10-6 we will use the “assumption” and 0.66-x = 0.66x2 = 0.660 x 4.76x10-6 x = [OH-] = 1.77 x 10-3 pOH = 2.75 and pH = 11.25

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12. You are given a weak acid with the formula HAn (122.1g/mol). When you make a solution of 0.500 g of HAn and 0.646 g Ca(An)2 in 0.500 L of water the pH is 5.92.(a) Calculate Ka for the acid.(b) If 2.0 mL of 1.0 M HCl are added to this solution what will be the pH?(c) How many grams of NaOH should be added to this solution to get a pH of 6.25.

Answer:(a) This is a mixture of a weak acid and its conjugate base

Ka = [An-][H+]/[HAn] [HAn]= 0.500 x 1mol/122.1g in 0.500L = 4.095 x 10-3 moles/0.500L [An-] = 0.646 x 1mol/282.1gx 2moles [An-]/mole of Ca salt in 0,500L

= 4.58 x 10-3 moles/0.500L[H+] = 10-5.92= 1.202 x 10-6

moles/L

And Ka = 1.34 x 10-6 = 4.58x 10-3x 1.202x10-6 /4.095x10-3

Could use the equation pH = pKa +log [A]/[HA] and will get the same number.

(b) have added 2.0 x10-3 L x 1mole/L = 2.0 x 10-3 moles HCl. This reacts with An- to yield HAn. Convert 2.0 x 10-3 moles of An- to HAn. New concentrations are

(4.095x 10-3 + 2.0 x 10-3)moles HAn/0.502L = 6.095x 10-3moles/0.502L

(4.58x 10-3 – 2.0 x 10-3)moles An-/0.502L = 2.58 x 10-3moles/0.502L

[H+] = Ka [HAn]/[An-] = 3.166 x 10-6

pH = 5.50

(c) For a pH =6.25, [H+] = 5.62 x 10-7M

the ratio[HAn]/[An-] = [H+]/Ka = 0.4197 = (4.095 x 10-3-x)/4.58 x 10-3 + x)

need to add enough OH- to adjust this ratio. x moles of OH- makes x moles of An- from x moles HAn.

Solve for x:

1.922 x 10-3 + 0.4197 x = 4.095x 10-3 –x

1.4197 x = 2.173x 10-3

x = 1.531 x 10-3 moles

need 1.531 x 10-3 moles x 40g NaOH/mole = 0.0612 g

13

13. (a) What is the minimum value of n for the following l values?l = 3 minimum n = 4

l = 2 minimum n = 3

(b) Circle (neatly!) the best single answer to the following question.

An atomic wavefunction represents:

i. the region of high probability for an electron around the nucleus of an atom

ii. the exact location of the electron

iii. the repulsion of all the electrons among themselves

iv. the region of high electron density for a covalent bond

v. an orbit that an electron follows around the nucleus of an atom

(c) How many orbitals are available with the following sets of quantum numbers?n = 4, l = 2, ml = 1 There is only one orbital (2 electrons)

n = 3, l = 3, ml = -1, ms= -1/2 zero orbitals, for n = 3 l cannot equal 3

n = 4, l = 3 ml =-3, -2, -1, 0, 1, 2, 3. There are 7 orbitals (they can hold 14 electrons)

14

14. For the reaction A → C, you are told that increasing the [A] by a factor of three leads to an increase in the reaction rate by a factor of 3. You are also given the following data.

[A] M time (min) 0.400 00.250 17.0

Propose a form for the rate law?What is the value of the rate constant, k? What is the half-life of the reaction? How long will it take for ¾ of the original concentration of A to be consumed?

Answer:

The tripling of the reaction rate with triple concentration indicate that the rate law is first order in A

Rate of reaction = k[A]

Using the integrated rate law provided:ln[A]t – ln[A]0 = -kt = -k(15min) = ln 0.250 – ln 0.400 = -0.4700

k = 0.0276 min-1

The half-life, t1/2 is the time when half of the original [A] has been consumed. Either use ln[A]t – ln[A]0 = -kt = -0.0276 min -1(t1/2) = ln 0.200 – ln 0.400 = -0.693 (t1/2) = -0.693/-0.0276 min -1 = 25.1 minor t1/2 = ln2/k = 0.693/0.0276(these are the same!)

Can either use the integrated rate law or realize that ¾ is 2 x t1/2

ln[A]t – ln[A]0 = -kt = -0.0276 min -1(t) = ln 0.1 – ln 0.400 = -1.386

50.2 min.

15