Test 2 related problems/worksheets

CHEM 1412 Buffers______

  1. Find the pH range of the following buffers: Calculate all pH to 2 decimal places.

Composition / Ka or Kb, indicate which / pKa / pKb / Effective pH range
1 / Benzoic acid/benzoate: C6H5CO2H/ C6H5CO2- / Ka = 6.3 x 10-5 / 4.20 / 9.80 / 3.20 - 5.20
2 / H2CO3/HCO3- / Ka = 4.2 x 10-7 / 6.38 / 7.62 / 5.38 – 7.38
3 / Nitrous acid/nitrite: HNO2/NO2- / Ka = 4.5 x 10-4 / 3.35 / 10.65 / 2.35-4.35
4 / H3PO4/H2PO4- / Ka1 = 7.5 x 10-3 / 2.12 / 11.88 / 1.12 – 3.12
5 / H2PO4-/HPO42- / Ka2 =6.2 x 10-8 / 7.21 / 6.79 / 6.21 – 8.21
6 / Aniline/anilium: C6H5NH2/ C6H5NH3+ / Kb = 4.2 x 10-10 / 4.62 / 9.38 / 3.62 – 5.62
7 / Ammonia/ammonium: NH3/NH4+ / Kb =1.8 x 10-5 / 9.26 / 4.74 / 8.26-10.26

B. Find the specific pH of the following buffers: pH = pKa + log([CB]/[CA])

Composition / [CB] / [CA] / [CB]/[CA] / pH / [H+]
8 / 0.10M C6H5CO2H/ 0.25 M C6H5CO2- / 0.25 / 0.10 / 2.5 / 4.60 / 2.5 x 10-5
9 / 0.69 g NaNO2 in 25 mL 0.10 M HNO2 / 0.40 / 0.10 / 4.0 / 3.95 / 1.1 x 10-4
10 / 0.36 g NaH2PO4 in 100 mL 0.020 M H3PO4 / 0.030 / 0.020 / 1.5 / 2.30 / 5.0 x 10-3
11 / 0.10 M NaH2PO4/ 0.25 M Na2HPO4 / 0.25 / 0.10 / 2.5 / 7.61 / 2.5 x 10-8
12 / 0.535 g NH4Cl in 50.mL 0.10M NH3 / 0.10 / 0.20 / 0.5 / 8.96 / 1.1 x 10-9

C. Find the ratio of [CB]/[CA] required to prepare pH 6.40 solution using the following buffers.

Buffer / FormulaCA / Formula CB / [CB]/[CA] / [CA]=? If [CB] =0.10M / Is buffer effective
[CB]/[CA]: 01-10
13 / H2CO3/ HCO3- / H2CO3 / HCO3- / 1.05 / 0.095 / Yes
14 / H2PO4-/ HPO42- / H2PO4- / HPO42- / 0.155 / 0.646 / yes
15 / C6H5NH2/ C6H5NH3+ / C6H5NH3+ / C6H5NH2 / 60.3 / 0.0017 / no

Q13. pH = pKa + log([CB]/[CA]) 6.40 = 6.38 + 0.02 = ,

,if, an effective buffer b/c [CB]/[CA] is between 0.1 and 10

CHEM 1412 Acids/Base Classification

Classify the following compounds and calculate the pH values & percent ionization (hydrolysis) of 0.10M solution. Fill answers in the blanks and show works on separate paper.

Classification / reaction: ionization
or hydrolysis / Ka/ or Kb / pK / pH / [OH-]

1

/

NaCl

/ salt of SB/SA / none / none / none / 7.00 / 1.00 E-7
2 / HClO4 / Strong acid / HClO4 H+ + ClO4- / none / none / 1.00 / 1.0 E-13
3 / NaNO2 / salt of SB/WA / NO2-+H2O HNO2+OH- / Kb= 2.2E-11 / pKb=10.66 / 8.17 / 1.48 E-6
4 / Fe(NO3)3 / salt of WB/SA
w/ small highly charged metal / Fe(H2O)63+ Fe(H2O)5(OH)2+ + H+ / Ka = 4.0 E-3 / pKa=2.40 / 1.74 / 5.5 E-11
5 / NaC6H5CO2 / salt of SB/WA / C6H5CO2-+H2O
OH-+HC6H5CO2 / Kb=1.6 E-10 / pKb=9.80 / 8.60 / 4.0 E-6
6 / NH4Br / salt of WB/SA / NH4+NH3+H+ / Ka = 5.6E-10 / pKa=9.26 / 5.13 / 1.35 E -9
7 / Ca(OH)2 / strong base / Ca(OH)2  Ca2+ + 2OH- / none / none / 13.3 / 0.20
8 / C6H5NH3Cl / salt of WB/SA / C6H5NH3  C6H5NH2+H+ / Ka=2.38E-5 / pKa=4.62 / 2.81 / 6.46 E -12
9 / Na2HPO4 / salt of SB/WA
amphiprotic / HPO42-+H2OOH-+ H2PO4-
HPO42 H+ + PO43 / Kb=1.6 E-7
Ka = 3.6E-13 / pKb=6.80 / 10.10 / 1.26 E -4
10 / NaHCO3 / salt of SB/WA
amphiprotic / HCO3-+H2OOH-+ CO32- (Kb1)
HCO3- H+ + CO32-(Ka2) / Kb1(2.38e-8) >Ka2(4.8e-11) / pKb = 7.62 / 9.69 / 4.95e-5
11 / NaHC2O4 / salt of SB/WA amphiprotic / HC2O4- H+ + C2O42-
HC2O4-+H2OOH-+ H2C2O4 / Ka=6.4 E-5
Kb = 1.7E-13 / pKa=4.19 / pH < 7
(2.60) / 3.95 E -12

Q9. Na2HPO4

Na2HPO4is salt of SB/WA in the H3PO4 family

The equilibrium constants for the first conjugate pair (H3PO4 and H2PO4- )are Ka1 and Kb1

The equilibrium constants for the second conjugate pair (H2PO4-and HPO42- )are Ka2 and Kb2

The equilibrium constants for the third conjugate pair (HPO42- and PO43-)are Ka3 and Kb3

Normally we would consider hydrolysis of HPO4-1 as CB of WA (H2PO4-)with a Kb

HPO42-+H2O OH-+ H2PO4- Kb=

But HPO42- is amphiprotic, it is CA of PO43-, with further ionization

HPO42-  H++ PO43- Ka = Ka3 = 3.6x10-13

In comparing the two possible reactions, Kb(1.6 x 10-7 ) Ka ( 3.6x 10-13), calculation of pH

should be based on Kb

RHPO42-+H2O OH-+ H2PO4-

E0.1-x x x

Q10. NaHCO3, NaHC2O4is salt of SB/WA in the H2CO3 family

The equilibrium constants for the first conjugate pair (H2CO3 and HCO3- )are Ka1 and Kb1

The equilibrium constants for the second conjugate pair (HCO3- and CO32- )are Ka2 and Kb2

Compare Kb1 (2.38e-8) and Ka2 (4.8e-11), we are going to use Kb1 to calculate pH

HCO3-+H2O OH-+ H2CO3- Kb1=

Q11. NaHC2O4, NaHC2O4is salt of SB/WA.

The equilibrium constants for the first conjugate pair (H2C2O4 and HC2O4- )are Ka1 and Kb1

The equilibrium constants for the second conjugate pair (HC2O4- and C2O42- )are Ka2 and Kb2

Normally we would consider hydrolysis of HPO4-1 as CB of WA, with a Kb

HC2O4-+H2O OH-+ H2C2O4- Kb1=

But HC2O4-- is amphiprotic, it is CA of C2O42-, with further ionization

HC2O4- H+ + C2O42- Ka = Ka2 = 6.4x10-5

In comparing the two possible reactions, Ka ( 6.4x 10-5) > Kb (1.7 x 10-13 ), calculation of pH should be

based on Ka

RHC2O4- H+ + C2O42-

E0.1-xxx

CHEM 1412 Acid-Base Reactions

Write the balance net ionic equation for the solutions in the far left column with HCl and NaOH.

CI: Common-ion effeect

solution / classification / Acid or base / reaction w/ HCl / reaction w/ NaOH
1 / HNO3 / SA / acid / N/A / H+ + OH- H2O
2 / CH3CO2H / WA / acid / NR, CI effect on ionization of CH3CO2H
CH3CO2HCH3CO2- +H+ / CH3CO2H+ OH- H2O + CH3CO2-
3 / C6H5NH2 / WB / base / C6H5NH2+ H+ C6H5NH3+ / NR, CI effect on ionization of C6H5NH2
C6H5NH2+ H2O  C6H5NH3+ +OH-
4 / NH3 / WB / base / NH3+ H+ NH4+ / NR, CI effect on ionization of NH3
NH3+ H2O  NH4+ +OH-
5 / H3PO4 / WA / acid / NR, CI effect on ionization of H3PO4
H3PO4 H2PO4- +H+ / H3PO4+ OH- H2O + H2PO4-
6 / KF / Salt of SB/WA / base / F-+ H+ HF / NR, CI effect on hydrolysis of F-
F-+ H2O  HF +OH-
7 / NaCN / Salt of SB/WA / base / Similar to Q6 / Similar to Q6
8 / NaH2PO4 / Salt of SB/WA / Amphi-protic / H2PO4-+ H+ H3PO4 / H2PO4- + OH- H2O + HPO42-
9 / C6H5NH3Cl / Salt of WB/SA / acid / NR, CI effect on hydrolysis of C6H5NH3+,
C6H5NH3+ C6H5NH2 + H+ / C6H5NH3++ OH- H2O + C6H5NH2
10 / NH4NO3 / Salt of WB/SA / acid / Similar to Q9 / Similar to Q9
11 / HNO2-NaNO2 / WA buffer / Conjugate pair / NO2--+ H+ HNO2 / HNO2+ OH- H2O+ NO2-
12 / NH3-NH4Cl / WB buffer / Conjugate pair / NH3+ H+ NH4+ / NH4+ + OH- H2O+ NH3

CHEM 1412 Worksheet Acid-Base reaction

1. 10.0mL 0.10M NaOH (SB)
+ 25.0mL 0.10M Hcl (SA)
. [ H+]=.1x25/35 = 0.0714M
[OH-]=.10x10/35 = 0.0286M
H+ + OH- H2O
.0714 .0286
[H+]=.0714-.0286 =.0428M pH=1.37 / 2. 10.0mL 0.10M NaOH (SB)
+ 25.0mL 0.10M HNO2 (WA)
.[HNO2] = 0.0714M [OH-]= 0.0286M
R HNO2+OH- NO2-+H2O
I .0714 .0286
C -.0286 -.0286 +.0286
E .0428  0 .0286
buffer, pH = 3.34 + log(CB/CA) = 3.17
3. 25.0mL 0.10M NaOH (SB)
+ 25.0mL 0.10M Hcl (SA)
. [H+]=0.050M [OH-]= 0.050M
H+ + OH- H2O
.050 .050
complete neutralization between SA & SB
[H+]=[OH-] =1.0x10-7M
pH=7.0 / 4. 25.0mL 0.10M NaOH (SB)
+ 25.0mL 0.10M HNO2 (WA)
[HNO2] = 0.050M [OH-]= 0.050M
neutralization HNO2+OH- NO2-+H2O
After neutralization, only 0.05 M NO2-hydrolysis
NO2-+H2O HNO2+OH-
Kb =2.2x10-11 =x2/(.05-x)
x = [OH-]=1.05x10-6M , pOH = 5.98,
pH=8.02
5. 50.0mL 0.10M NaOH (SB)
+ 25.0mL 0.10M Hcl (SA)
[H+]=0.033M [OH-]= 0.067M
H+ + OH- H2O
After neutralization,
[OH-] =.034 pH=12.53 / 6. 50.0mL 0.10M NaOH (SB)
+ 25.0mL 0.10M HNO2 (WA)
[HNO2] = 0.033M [OH-]= 0.067M
neutralization HNO2+OH- NO2-+H2O
.After neutralization,
[OH-]=.034M, pOH=1.47 , pH=12.53
7. 25.0mL 0.10M NH3(WB)
+ 25.0mL 0.10M Hcl (SA)
[H+]=0.050M [NH3]= 0.050M
neutralization
R H+ + NH3  NH4+
I 0.050 0.050 0
C -0.050 -0.050 +0.050
E ~0 ~0 0.050
Hydrolysis of NH4+ Ka = x2/(0.05-x)
x = [H+]=5.3x10-6M , pH = 5.28
x = [H+] = 5.3 x 10-6 M pH = 5.28 / 8. 50.0mL 0.10M NH3(WB)
+ 25.0mL 0.10M Hcl (SA)
[H+]=0.033M [NH3]= 0.067M
R H+ + NH3  NH4+
I 0.033 0.067 0
C -0.033 -0.033 +0.033
E ~0 0.034 0.033
A buffer combination
pH = pKa + log(0.034/0.033) = 9.27

SAMPLE TEST 2

2. Which of the following pair can be used to prepare a buffer? Why?

a. 0.1 M HCl and 0.1 M NaOHNo, SA and SB

b. 0.1 M NaOH and 0.1 M NaC2H3O2 No, SB (NaOH) and WB (NaC2H3O2 as salt of SB/WA)

c. 0.1 M HC2H3O2and 0.1 M NaC2H3O2Yes, Conjgate pair , WA (HC2H3O2) and CB (C2H3O2-)

d. 0.1 M HCl and 0.1 M NH3No, SA and WB in equal concentration

e. 0.2 M NaOH and 0.1 M HC2H3O2No, [SB] > [WA]

f. 0.2 M HCl and 0.1 M NH3No, [SA] > [WB]

g 0.1 M NaOH and 0.1 M HC2H3O2 No, [SB] = [WA]

h. 0.1 M HCl and 0.1 M HC2H3O2No, two acids, SA (HCl) and WA (HC2H3O2)

i. 0.1 M NaOH and 0.2 M HC2H3O2 Yes, [WA] > [SB]

j. 0.1 M HCl and 0.2 M NH3Yes, [WB] > [SA]

k. 0.1 M HCl and 0.1 M NaClNo, SA and its CB

5. Which of the following solution can neutralize the most NaOH? The least NaOH? Explain.

a. 50.0mL of 0.10M HCl b50.0mL of 0.10M HC2H3O2 . c. 50.0mL of 0.10M H2SO4

Ans: the amount of NaOH neutralized by acid depends on the moles of available acidic H (equivalent)

= equivalents of acid = MxVx(# acidicH)

a. 50.0mL of 0.10M HCl = (0.10M )(0.050L) (1) = 0.0050 equi.

b. 50.0mL of 0.10M HC2H3O2 = (0.10M )(0.050L) (1) = 0.0050 equi

c. 50.0mL of 0.10M H2SO4= (0.10M )(0.050L) (2) = 0.010 equi

c > a = b

Q6. a. If 0.50 mL of 3.0 M HCl added to 100.00 mL DI-water

HCl will be diluted. Use M1V1 = M2V2

[H+] = (3.0M)(0.50 mL) /(100.5 mL) = 0.015 M; pH = 1.83

Q7. a. pH of buffer , use pH = pKa + log[CB/CA] = 4.74 + log (0.2/0.1) = 5.04

b. HCl will be diluted by buffer solution:[H+] = (3.0M)(0.50 mL) /(100.5 mL) = 0.015 M

Next look at reaction of buffer w/ HCl: It is the CB in the buffer to react w/ HCl

R C2H3O2- + H+  HC2H3O2

I 0.20 0.015 0.10

C -0.015 -0.015 +0.015

E 0.185  0 0.115 Still a buffer

c. pH after NaOH is added:

Work out neutralization reaction of buffer w/ NaOH, use RICE for calculation (like in 7b.)

R HC2H3O2 + OH-C2H3O2- + H2O

I0.10 0.0150.20

C-0.015 -0.015 +0.015

E0.085  00.215

Q8. Hint: Consider common-ion effect of [H+] on ionization of acetic acid

Q9. Check problem set Q10.

T2 worksheet Su07 key v2.doc110/9/201811:51:03 AM