8
Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol 1.00 mol L x L x = 0.075L H + + Cl - + Na + + OH - H 2 O + Na + + Cl - NaCl (s) [Na + ]= 0.075 mol .150 L + .075 L = 0.333 M = [Cl - ] base strong 150 mL0.500 M HCl = mol H + 0.500 mol = 0.15 L acid strong =

Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH - H2OH2O+ Na + + Cl - NaCl

Embed Size (px)

Citation preview

Page 1: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

Saltsproduct of neutralization reaction

1.00 M NaOH

mol OH-

L0.075 mol1.00 mol

Lx L

x = 0.075L

H+ + Cl- + Na+ + OH- H2O + Na+ + Cl-

NaCl (s)[Na+] = 0.075 mol

.150 L + .075 L

= 0.333 M = [Cl-]

basestrong150 mL 0.500 M HCl

= mol H+

0.500 mol= 0.15 L

acidstrong

=

Page 2: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

NaCl (s) HClNaOH

conjugate baseconjugate acid

Cl-Na+

very weak very weak

Cl- + H2O no reactionNa ++ H2O no reaction

Saltsproduct of neutralization reaction

basestrong acidstrong

Li+

K+

Rb+

Ca2+

Sr2+

Ba2+

negligible acidity

Br-

I-

NO3-

ClO4-

negligible basicity

Page 3: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

Saltsproduct of neutralization reaction

base acidstrong weak

1.00 M NaOH

mol OH-

L0.075 mol1.00 mol

Lx L

150 mL 0.500 M HF

= mol H+

0.500 mol= 0.15 L=

x = 0.075L

H+ + F- + Na+ + OH- H2O + Na+ + F-

NaF (s)

HF

Ka = 1.5 x 10-11

Page 4: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

Saltsproduct of neutralization reaction

base acidstrong weakNaF (s)NaOH HF

conjugate baseconjugate acidvery weak strong

Na ++ H2O no reaction F- + H2O HF + OH-

Ka = 1.5 x 10-11

Ka x Kb = Kw

Kb =6.7 x 10-4hydrolysis

“breaking water”

Na+ F-

basic solution

=[HF] [OH-]

[F-]

Page 5: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

Saltsproduct of neutralization reaction

base acidweak strong

HClO4CH3NH2

CH3NH2+ H2O CH3NH3++ OH- + H+ + ClO4

-

CH3NH3ClO4 (s)

H2O

CH3NH3+ ClO4

-

conjugate baseweakconjugate acidstrong

Kb = 4.4 x 10-4

Ka = 2.3 x 10-11

CH3NH3+ + H2O CH3NH2 + H3O+

acidic solutionhydrolysis

Page 6: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

Saltsproduct of neutralization reaction

base acidweak weak

NH3HNO2

NH4NO2

NH4+

NO2-

NH3

HNO2

+ H2O + H3O+

+ H2O + OH-

Ka =

Kb =

5.7 x 10-10

1.4 x 10-11

acidic solution

Page 7: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

List the following species in order of

C5H5NHCl

K2CO3

LiNO3

NaHCO3

cation anion pH

C2H5NH+ Cl-

K+ CO32-

Li+ NO3-

Na+ HCO3-

w.b.

s.b.

s.b.

s.b.

s.a.

s.a.

w.a.

w.a.

acidic

basic

neutral

basic

I

II

III

IV

0.20 M

increasing pH

Page 8: Salts product of neutralization reaction 1.00 M NaOH mol OH - L 0.075 mol1.00 mol L x L x = 0.075L H+H+ + Cl - + Na + + OH -  H2OH2O+ Na + + Cl - NaCl

Calculate the pH of a

C2H5NH3 NO3+ -

base acid

C2H5NH3+ + H2O C2H5NH2+ H3O+

Kb = 5.6 x 10-4

Ka x Kb = Kw = 1.0 x 10-14 Ka = 1.8 x 10-11

1.8 x 10-11 = [H3O+] [C2H5NH2]

[C2H5NH3+]

= x2

0.1

x = 1.34 x 10-6 pH = 5.87

0.10 M C2H5NH3NO3 solution

weak strong