PH DETERMINATION

J.R.A. Ibale INSTITUTE OF CHEMISTRY, COLLEGE OF SCIENCE

UNIVERSITY OF THE PHILIPPINES, DILIMAN QUEZON CITY, PHILIPPINES DATE PERFORMED: JANUARY 25, 2013 INSTRUCTOR’S NAME: MARO PEÑA

REFERENCES

Petrucci, R.H., et al. General Chemistry, 10th Ed . 2010.

APPENDICES CALCULATIONS

A. pH Measurement of Solutions 1. Distilled water

pH: 6.01 pOH: 14 – pH = 7.99 [H3O+] = 10-pH = 9.77 x 10-7 M [OH-] = 10-pOH = 1.02 x 10-8 M

2. 0.10 M H3PO4 pH: 1.96 pOH: 14 – pH = 12.04 [H3O+] = 10-pH = 0.011 M [OH-] = 10-pOH = 9.12 x 10-13 M

3. 0.10 M NaH2PO4 pH: 4.58 pOH: 14 – pH = 9.42 [H3O+] = 10-pH = 2.63 x 10-5 M [OH-] = 10-pOH = 3.80 x 10-10 M

4. 0.10 M Na2HPO4 pH: 9.26 pOH: 14 – pH = 4.74 [H3O+] =10-pH = 5.50 x 10-10 M [OH-] = 10-pOH = 1.82 x 10-5 M

5. 0.10 M Na3PO4 pH: 12.28 pOH: 14 – pH = 1.72 [H3O+] =10-pH = 5.25 x 10-13 M [OH-] = 10-pOH = 0.02 M

6. 0.10 M NaHCO3 pH: 9.36 pOH: 14 – pH = 4.64 [H3O+] =10-pH = 4.37 x 10-10 M [OH-] = 10-pOH = 2.29 x 10-5 M

7. 1.0 x 10-8 M HCl pH: 7.15 pOH: 14 – pH = 6.85 [H3O+] =10-pH = 7.08 x 10-8 M [OH-] = 10-pOH = 1.41 x 10-7 M

8. 1.0 x 10-8 M NaOH pH: 7.03 pOH: 14 – pH = 6.97 [H3O+] =10-pH = 9.33 x 10-8 M [OH-] = 10-pOH = 1.07 x 10-7 M

9. 0.10 M ethylenediamine pH: 10.23 pOH: 14 – pH = 3.77 [H3O+] =10-pH = 5.89 x 10-11 M [OH-] = 10-pOH = 1.70 x 10-4 M

10. 0.10 M NH4CH3COO pH: 7.10 pOH: 14 – pH = 6.90 [H3O+] =10-pH = 7.94 x 10-8 M [OH-] = 10-pOH = 1.26 x 10-7 M

B. pH and Degree of Ionization of Solutions

Note: SA (Strong Acids—dissociate completely) WA (Weak Acid) SB (Strong Bases—dissociate completely) WB (Weak Base)

α = %ionization =

x 100%

1. 0.10 M HCl SA

HCL ⇌ H+ + Cl-

pH = -log(0.10 M) = 1 α = 100%

2. O.10 M HNO3 SB HNO3 ⇌ H+ + NO3

-

pH = -log(0.10 M) = 1 α = 100%

3. O.10 M CH3COOH WA Table 1. ICE Table for CH3COOH Dissociation

CH3COOH(aq) + H2O(aq) ⇌ CH3COO-(aq) + H3O+(aq)

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Ka = [

][ ]

= 1.8 x 10-5 =

x = √ ( ) = 1.341640786 M = [H3O+]

pH = -log[H3O+] = 2.87

α = [

]

= 1.31%

4. 0.10 M H3PO4 WA

Table 1. ICE Table for CH3COOH Dissociation

H3PO4(aq) + H2O(aq) ⇌ H2PO4-(aq) + H3O+(aq)

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Ka =

= 7.1 x 10-3

x2 + 7.1 x 10-3(x) - 7.1 x 10-3(0.10 M) = 0 x = 0.023331266 M = [H3O+] pH = -log(0.023331266 M) = 1.63

5. 0.10 M NaOH SB NaOH(aq) Na+(aq) + OH-(aq)

pOH = -log [OH-] = 1 pH = 14 − pOH = 13 α = 100%

6. 0.10 M KOH SB KOH(aq) K+(aq) + OH-(aq)

pOH = -log [OH-] = 1 pH = 14 − pOH = 13 α = 100%

7. 0.10 M NH3 WB Table 2. ICE Table for CH3COOH Dissociation

NH3 + H2O ⇄ NH4+ + OH-

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Kb = 1.8 x 10-5

[OH-] = √ ( ) = 1.341640787 x 10-3 M pOH = - log [OH-] = 2.87 pH = 14 − pOH = 11.13

α= [ ]

[ ] = 1.31%

8. 0.10 M NaCl Neutral Salt

Note: Salt of HCl (SA) and NaOH (SB) salt of neutralization rxn H+ + OH- ⇌ H2O(l) pH = 7.00

9. 0.10 M NH4CH3COO Neutral Salt Note: Salt of NH3 (Kb = 1.8 x 10-5) and CH3COOH (Ka = 1.8 x 10-5) salt of neutralization rxn H+ + OH- ⇌ H2O(l) pH = 7.00

10. 0.10 M NH4Cl Acidic Salt Table 3. ICE Table for Hydrolysis of NH4+

NH4+ + H2O ⇄ NH3 + H3O+

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Kb = 1.8 x 10-5

Ka =

=

= 5.56 x 10-10 =

[H3O+] = 7.456540753 x 10-6 M pH = -log (7.456540753 x 10-6 M) = 5.13

α= [ ]

[ ] = 0.007%

11. 0.10 M NaH2PO4 Acidic Salt (of a Polyprotic Acid; Ampholyte)

H2PO4-(aq) + H2O ⇌ H3PO4(aq) + OH-(aq)

Ka1 = 7.1 x 10-3 (H3PO4) Ka2 = 6.3 x 10-8 (H2PO4-)

[H3O+] = √ = 2.1149 x 10-5 M

pH = - log [H3O+] = 4.67

12. 0.10 M Na2HPO4 Basic Salt (of a Polyprotic Acid; Ampholyte) HPO42-(aq) + H2O ⇌ H2PO4-(aq) + OH-(aq)

Ka1 = 6.3 x 10-8 (H2PO4-) Ka2 = 4.2 x 10-13 (HPO42-) Kb1 = 1.5873 x 10-7 Kb2 = 0.02381

[OH-] = √ = 6.1476 x 10-5 M

pOH = -log [OH-] = 4.21 pH = 9.79

13. 0.10 M Na3PO4 Basic Salt

Table 4. ICE Table for Hydrolysis of PO43-

PO43- + H2O ⇄ HPO42- + OH-

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Ka = 4.2 x 10-13

Kb =

=

= 0.02380952381 =

x2 + Kbx – Kb(0.10 M) = 0 x = 0.038321483 = [OH-] pOH = -log[OH-] = 1.42 pH = 14 − pOH = 12.58

14. 0.10 NaCH3COO Acidic Salt

Table 4. ICE Table for Hydrolysis of CH3COO-

CH3COO- + H2O ⇄ CH3COOH + OH-

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Ka = 1.8 x 10-5

Kb =

=

= 5.56 x 10-10 =

x = √ ( ) = 7.453559925 x 10-6 M = [OH-]

pOH = -log[OH-] = 5.13 pH = 14 − pOH = 8.87

α = [ ]

[ ] = 0.007%

15. 0.10 M NaHCO3 Basic Salt (of a Polyprotic Acid; Ampholyte)

HCO3-(aq) + H2O ⇄ H2CO3(aq) + OH-(aq)

Ka1 = 4.4 x 10-7 (H2CO3) Ka2 = 4.7 x 10-11 (HCO3-)

[H+] = √ = 4.5475268 x 10-9 M

pH = -log[H+] = 8.34

16. 1.0 x 10-8 HCl Higly Diluted Strong Acid

Table 4. ICE Table for Autoionization of H2O

H2O ⇄ H3O+(aq) + OH-(aq)

Initial 1.0 x 10-8 --

Change +x +x

Equilibrium 1.0 x 10-8+x +x

Kw = (1.0 x 10-8 + x)x = 1.0 x 10-14 x2 + 1.0 x 10-8 x – Kw = 0 x = 9.512492197 x 10-8 M [H3O+] = 1.0 x 10-8 + x = 1.05124922 x 10-7 M pH = -log[H3O+] = 6.98

17. 1.0 x 10-8 M NaOH Highly Diluted Strong Base Table 5. ICE Table for Autoionization of H2O

H2O ⇄ H3O+(aq) + OH-

(aq)

Initial -- 1.0 x 10-8

Change +x +x

Equilibrium +x 1.0 x 10-8+x

Kw = (1.0 x 10-8 + x)x = 1.0 x 10-14 x2 + 1.0 x 10-8 x – Kw = 0 x = 9.512492197 x 10-8 M = [H3O+] pH = -log[H3O+] = 7.02

18. 0.10 M ethylenediamine (H2NC2H4NH2) Weak Base (Organic; Amine) H2NC2H4NH2 + H2O ⇄ H3NC2H4NH2 + OH- Kb1 = 8.0 x 10-5

H3NC2H4NH2 + H2O ⇄ H3NC2H4NH3 + OH- Kb2 = 9.0 x 10-8

Table 4. ICE Table for Hydrolysis of CH3COO-

H2NC2H4NH2 + H2O ⇄ H3NC2H4NH2 + OH-

Initial 0.1 -- --

Change -x +x +x

Equilibrium 0.1-x +x +x

Kb1 =

x = √ ( ) = 2.828427125 x 10-3 M = [OH-]

pH = 14 − -log[OH-] = 11.45

α = [ ]

[ ] = 2.8% (In this calculation, it has been initially assumed that all

H3O+ came from the first ionization of ethelynediamine)