Salt Ion(s) that will hydrolyze Spectator Ions Universal indicator color Measure
ID: 1004769 • Letter: S
Question
Salt
Ion(s) that will hydrolyze
Spectator Ions
Universal indicator color
Measured pH (sensor)
[H+] From pH
[OH-] From pH
Net Ionic equation
Ka/Kb expression
Ka/Kb value (use ICE tables)
Tap
Water
------
------
Gn
8.46
3.46E-9
2.88E-6
------
------
------
D.I.
H2O
------
------
Or
5.00
1.00
E-5
1.00E-9
------
------
------
NaCl
------
Na+,Cl-
Y/Or
6.40
3.89E-7
2.51E-8
------
------
------
NaC2H3O2
C2H3O2-
Na+
Y/Gn
7.70
2.00
E-8
5.01E-7
C2H3O2-+H2O <->HC2H3O2+OH-
Kb= 2.51E-12
NH4Cl
NH4+
Cl-
Y
5.84
1.45E-6
6.92E-9
NH4++H2O <-> NH4OH + H+
Ka = 2.10E-11
ZnCl2
Zn2+
2Cl-
Y
4.49
3.24E-5
3.09E-10
Zn2++H2O <-> ZnOH+ + H+
Ka = 1.05E-8
KAl(SO4)2
Al3+
K-2SO4-
R
3.03
9.33E-4
1.01E-11
Al3++H2O <-> AlOH2++H+
Ka = 8.79E-6
Na2CO3
CO32-
2Na+
Dark Blue
11.2
6.31E-12
1.58E-3
CO32-+H2O <-> HCO3-+OH-
Kb = 2.54E-5
a)
How would the pH of 21.5 mL of pure water change upon addition of 0.25 mL of 6.0 M HCl? Upon addition of 0.25 mL 6.0 M NaOH? Compare these values to the values obtained for your buffer upon similar additions
Salt
Ion(s) that will hydrolyze
Spectator Ions
Universal indicator color
Measured pH (sensor)
[H+] From pH
[OH-] From pH
Net Ionic equation
Ka/Kb expression
Ka/Kb value (use ICE tables)
Tap
Water
------
------
Gn
8.46
3.46E-9
2.88E-6
------
------
------
D.I.
H2O
------
------
Or
5.00
1.00
E-5
1.00E-9
------
------
------
NaCl
------
Na+,Cl-
Y/Or
6.40
3.89E-7
2.51E-8
------
------
------
NaC2H3O2
C2H3O2-
Na+
Y/Gn
7.70
2.00
E-8
5.01E-7
C2H3O2-+H2O <->HC2H3O2+OH-
Kb= 2.51E-12
NH4Cl
NH4+
Cl-
Y
5.84
1.45E-6
6.92E-9
NH4++H2O <-> NH4OH + H+
Ka = 2.10E-11
ZnCl2
Zn2+
2Cl-
Y
4.49
3.24E-5
3.09E-10
Zn2++H2O <-> ZnOH+ + H+
Ka = 1.05E-8
KAl(SO4)2
Al3+
K-2SO4-
R
3.03
9.33E-4
1.01E-11
Al3++H2O <-> AlOH2++H+
Ka = 8.79E-6
Na2CO3
CO32-
2Na+
Dark Blue
11.2
6.31E-12
1.58E-3
CO32-+H2O <-> HCO3-+OH-
Kb = 2.54E-5
a)
How would the pH of 21.5 mL of pure water change upon addition of 0.25 mL of 6.0 M HCl? Upon addition of 0.25 mL 6.0 M NaOH? Compare these values to the values obtained for your buffer upon similar additions
Explanation / Answer
a)
How would the pH of 21.5 mL of pure water change upon addition of 0.25 mL of 6.0 M HCl?
0.25 mL of 6.0 M HCl solution = 6.0 mol/L * 0.25 mL * 1 L/1000 mL = 0.0015 mol H+
[H+] from pure H2O = 1.0*10-14 mol/L
We have 21.5 mL pure water
Moles of H+ from 21.5 mL pure water = 1.0*10-14 mol/L *21.5 mL *1 L /1000 mL = 2.15*10-16 mol H+
Total number of moles of H+ = 0.0015 mol H+ + 2.15*10-16 mol H+ = 0.0015 mol H+
[H+] = 0.0015 mol H+ / (0.0215 + 0.00025) L = 0.06896 M
pH = - log [H+]= - log (0.06896 ) = 1.16
---------------------------------------------------------------------------------------------------------------------------------------------------
Upon addition of 0.25 mL 6.0 M NaOH?
Pure water [OH-] =1.0*10-14 M
Volume of pure water = 21.5 mL = 0.0215 L
Moles of OH- from pure water = 1.0*10-14 mol/L * 0.0215 L = 2.15*10-16 mol OH-
Moles of OH- from 0.25 mL 6.0 M NaOH = 6.0 mol/L * 0.25 mL * 1 L/1000 mL = 0.0015 mol OH-
Total number of moles of OH- = 2.15*10-16 mol OH- + 0.0015 mol OH-= 0.0015 mol OH-
[OH-] = 0.0015 mol OH- / (0.0215 + 0.00025) L = 0.06896 M OH-
pOH = - log [OH-]= - log (0.06896 ) = 1.16
pH + pOH = 14.00 = pH + pOH = 1.16
pH = 14.00 - 1.16 = 12.84
---------------------------------------------------------------------------------------------------------------------------------------------------
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.