CHEM 1412 Weak Acids/weak Bases Ionization worksheet Name: ______

Dr. Ya-Ping Huang Fill the blanks. Show your work on separate sheet of paper. Synonym: ______

Compound / Initial Conc. / % ionization / [H+] / [conj.base]eq
[conj.acid]eq / pH / pOH / K
1 / unknown
acid HX / 0.10M / . 4.0 x 10-1 / 4.0 x 10-4 M / 4 x 10-3 / 3.40 / 10.60 / 1.6 x10-6
2 / unknown
acid HX / 0.25M / 0.85% / 2.12 x 10-3 M / 8.55 x 10-3 / 2.67 / 11.33 / 1.8x10-5
3 / unknown
base B / 1.5 x 10-4M / 16.7% / 4.0 x 10-10M / 5.0 / 9.4 / 4.6 / Kb = 5.0 x 10-6
4 / unknown
acid HX / 6.56x10-3 / 15.2 % / 1.0x 10-3 / 0.18 / 3.0 / 11.0 / Ka = 1.8 x10-4

CHEM 1412 Assignment Buffers Synonym______

Find the pH range of the following buffers:

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

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

Composition / Ratio [CB]/[CA] / pH / [H+]
0.10M C6H5CO2H/ 0.25 M C6H5CO2- / 2.5 / 4.60
0.01 M H2CO3/0.02M HCO3- / 2.0 / 6.68
0.690 g NaNO2 in 25 mL 0.10 M HNO2 / 4.0 / 3.95
0.36 g NaH2PO4 in 100 mL 0.020 M H3PO4 / 1.5 / 2.30
0.10 M NaH2PO4/ 0.25 M Na2HPO4- / 2.5 / 7.61
0.535 g NH4Cl in 50.mL 0.10M NH3 / 0.5 / 8.96
Find the ratio of [CB]/[CA] required to prepare pH 6.20 solution using the following buffers. Show the formulas of conj base and conj. acid ( i.e. [C2H3O2-]/[HC2H3O2] = 2.5), also commenting on the effectiveness of each buffer
H2CO3/ HCO3- / ([HCO3-]/[ H2CO3])= 10(E-0.18)=0.66
HNO2/NO2- / [NO2-]/[ HNO2] = 10(E2.875) = 708
H2PO4-/ HPO42- / [HPO42-]/[H2PO4-]=10(E-1.01)= 0.098
C6H5NH2/ C6H5NH3+ / [C6H5NH2 ]/[ C6H5NH3+] =10(E1.58)= 38.

CHEM 1412 Acids/Base Classification Name :

Classify the following compounds and calculate the pH values & percent ionization (hydrolysis) of 0.10M solution.

For pH *estimate, please indicate if it’s > 7, = 7 or < 7. Fill answers in the blanks and show works on separate paper.

Classification / reaction: ionization or hydrolysis / Ka/ or Kb / % ionization(acid, base)
or % hydrolysis (salt) / pH / [OH-]

1

/

NaCl

/ salt of SB/SA / none / none / 0 % hydrolysis / 7.0
2 / HClO4 / Strong acid / HClO4® H+ + ClO4- / none / 100 % ionization / 1.0
3 / NaNO2 / salt of SB/WA / NO2-+H2OÛHNO2+OH- / Kb= 2.2x10-11 / 0.00148% hydrolysis / 8.17
4 / Fe(NO3)3 / salt of WB/SA / Fe(H2O)63+Û Fe(H2O)5(OH)2+ + H+ / Ka = 4.0 x 10-3 / 18.1% / 1.74
5 / NaC6H5CO2 / salt of SB/WA / C6H5CO2-+H2OÛ OH-+HC6H5CO2 / Kb=1.6 x 10-10 / 4.0 X 10-3 % hydrolysis / 8.60
6 / NH4Br / salt of WB/SA / NH4+ÛNH3+H+ / Ka = 5.6x 10-10 / 0.00748% hydrolysis / 5.13
7 / Ca(OH)2 / strong base / Ca(OH)2 ® Ca2+ + 2OH- / none / 100 % ionization (dissociation) / 13.3
8 / C6H5NH3Cl / salt of WB/SA / C6H5NH3 Û C6H5NH2+H+ / Ka=2.38x10-5 / 1.54% hydrolysis / 2.81
9 / Na2HPO4 / salt of SB/WA
amphiprotic / HPO42-+H2OÛ OH-+ H2PO4-
HPO42Û H+ + PO43 / Kb=1.6 x 10-7 Ka = 3.6x 10-13 / Skip / Estimate
> 7 / Skip
10 / NaHC2O4 / salt of SB/WA amphiprotic / HC2O4-Û H+ + C2O42-
HC2O4-+H2OÛ OH-+ H2C2O4 / Ka=6.4 x 10-5 Kb = 1.7x 10-13 / Skip / Estimate
< 7 / Skip

CHEM 1412 Assignment ACID-BASE REXN KEYS Dr. Ya-Ping Huang

Calculate the pH value for the following solutions: You need to show more details than the one shown below

1. 10.0mL 0.10M NaOH +25.0mL 0.10M Hcl
neutralization:
H+ + OH- ® H2O
[H+]=0.0428M pH=1.37 / 2. 10.0mL 0.10M NaOH +25.0mL 0.10M HNO2
neutralization:
HNO2+ + OH- ® H2O + NO2-
Then pH = pKa + log(CB/CA)
[H+] =6.73x10-4, pH=3.17
3. 25.0mL 0.10M NaOH +25.0mL 0.10M Hcl
neutralization:
H+ + OH- ® H2O
complete neutralization of SA & SB
[H+]=[OH-] =1.0x10-7M pH=7.0 / 4. 25.0mL 0.10M NaOH +25.0mL 0.10M HNO2
neutralization:
HNO2 + OH- ® H2O + NO2-
Then hydrolysis:
NO2- + H2O Û HNO2 + OH-
. [OH-]=1.05x10-6M pOH=5.98 , pH=8.02
5. 50.0mL 0.10M NaOH +25.0mL 0.10M Hcl
neutralization:
H+ + OH- ® H2O
[OH-] = 0.034 pOH =1.47 pH = 12.53 / 6. 50.0mL 0.10M NaOH +25.0mL 0.10M HNO2
neutralization:
HNO2 + OH- ® H2O + NO2-
then [OH-]=.034M pOH=1.47 , pH = 12.53
7. 25.0mL 0.10M NH3 +25.0mL 0.10M Hcl
neutralization:
NH3+ H+® NH4+
Then hydrolysis: NH4+Û NH3+ H+
[H+]=5.3x10-6 M, pH = 5.28 / 8. 50.0mL 0.10M NH3 +25.0mL 0.10M Hcl
neutralization:
NH3+ H+® NH4+
Then buffer,
pH = 9.27

Sample Test 2 Problem #6. Titration curves:

When you draw the titration curves, pay attention to the details,

a.  the steep change of pH around equivalence point,

b.  the flat first segments (slow change of pH) when SA is titrated w/ SB or when SB is titrated w/ SA.

c.  the initial relatively large change in pH when WA is titrated w/ SB or when WB is titrated w/ SA.

d.  Draw the curve as ideal curve rather than just connect 4 dots (4 points of a-d). Refer to titration handout fro details

a. 25.0 mL of 0.1 M HCl titrated w/ 50.0 mL 0.10 M NaOH

Volume at equivalence point: [acid](Vacid)(# of acidic H) = [base](Vbase)(# of OH)
(0.10M)(25.0x10-3L)(1) = (0.10M)(VNaOH)(1)
VNaOH = 25.0 mL(25.0x10-3L )
Volume at half-neutralization point = ½ (Volume at equivalence point)
= ½(25.0 mL) = 12.5 mL

1. pH at the starting point ( w/ no second reagent))

[H+] = [HCl] = 0.10 M , pH = -log(0.10) = 1.0

2. Volume of the second reagent and pH at the half-neutralization point

Vol = 12.5 mL of NaOH

[H+] = [HCl] = (0.10 M x 25.0 mL)/37.5 mL = 0.0667 M

[OH-] = [NaOH] = (0.10 M x 12.5 mL)/37.5 mL = 0.0333 M

Neutralization of SA(HCl) and SB(NaOH)

R H+ + OH- ® H2O

I 0.0667 0.0333

C -0.0333 -0.0333

E 0.0334 ~ 0

Leftover SA [H+] will determine pH, pH = -log(0.0334) = 1.48

3. Volume of the second reagent and pH at the equivalent point

Vol = 25.0 mL of NaOH

[H+] = [HCl] = (0.10 M x 25.0 mL)/50.0 mL = 0.050 M

[OH-] = [NaOH] = (0.10 M x 25.0 mL)/50.0 mL = 0.050 M

Complete neutralization of SA(HCl) and SB(NaOH)

R H+ + OH- ® H2O

I 0.050 0.050

C -0.050 -0.050

E ~ 0 ~ 0

SA (HCl) is completely neutralized by SB(NaOH), solution is neutral, pH = 7.0

4. pH value when 50 mL of second reagents has been added, total volume = 75.0 mL

[H+] = [HCl] = (0.10 M x 25.0 mL)/75 mL = 0.0333 M

[OH-] = [NaOH] = (0.10 M x 50.0 mL)/75 mL = 0.0667 M

Neutralization of SA(HCl) and SB(NaOH)

R H+ + OH- ® H2O

I 0.0333 0.0667

C -0.0333 -0.0333

E ~ 0 0.0334

Leftover SB [OH-] will determine pH,

pOH = -log(0.0334) = 1.48 , pH = 14 –1.48 = 12.52

5. For SA and SB, no Ka or Kb information

b. 25.0 mL of 0.1 M NaOH titrated w/ 50.0 mL 0.10 M HCl