AcidsLesson 14

AmphiproticIons

Acids or Bases?

Kwantlen Science Challenge:

Kwantlen Science Challenge:

Kwantlen Science Challenge:

Amphiprotic Ions H3PO4 AcidH2PO4

- AmphiproticHPO4

2- AmphiproticPO4

3- Base Amphiprotic ions have a Ka and a Kb.

If the Ka > Kb, then the ion is an acid in water.

If the Kb > Ka, then the ion is a basein water.

Is H2PO4- acid or basic

acid

1. Is H2PO4- an acid or base when in water? Write the predominate hydrolysis

equation for this ion.  

Ka (H2PO4-) = 6.2 x 10-8

Kb (H2PO4-) = Kw

Ka(H3PO4)

= 1 x 10-14

7.5 x 10-3

= 1.3 x 10-12

Ka > Kb Acid

H2PO4- + H2O ⇄

HPO42- + H3O+

Is HPO42- acid or basic

basic

2. Is HPO42- an acid or base when in water? Write the hydrolysis equation for this

ion.  

Ka (HPO42-) = 2.2 x 10-13

 Kb (HPO4

2-) = KwKa(H2PO4

-)

= 1 x 10-14

6.2 x 10-8

= 1.6 x 10-7

Kb > Ka base

HPO42- + H2O ⇄

H2PO4- + OH-

3. Show by calculation if NH4F is an acid or base when in water.

NH4F → NH4+ + F-

acid baseBoth have weak parents- neither will cross off!What will the salt be?

Ka(NH4+) = 5.6 x 10-10

Kb (F-) = = KwKa(HF)

= 1 x 10-14

3.5 x 10-4

= 2.9 x 10-11

Ka > Kb acid

4. Show by calculation if NH4CH3COO is an acid or base when in water.

NH4CH3COO → NH4+ + CH3COO-

acid base

Ka(NH4+) = 5.6 x 10-10

Kb (CH3COO-) = = KwKa(CH3COOH)

= 1 x 10-14

1.8 x 10-5

= 5.6 x 10-10

Ka = Kb neutral pH = 7.00

5. Calculate the volume of 0.100 M H2CO3 required to neutralize 25.0 mL of 0.200 M KOH.

1H2CO3 + 2KOH → K2CO3 + 2HOH? L 0.0250 L0.100 M 0.200 M

5. Calculate the volume of 0.100 M H2CO3 required to neutralize 25.0 mL of 0.200 M KOH.

1H2CO3 + 2KOH → K2CO3 + 2HOH? L 0.0250 L0.100 M 0.200 M

0.0250 L KOH

5. Calculate the volume of 0.100 M H2CO3 required to neutralize 25.0 mL of 0.200 M KOH.

1H2CO3 + 2KOH → K2CO3 + 2HOH? L 0.0250 L0.100 M 0.200 M

0.0250 L KOH x 0.200 mol L

5. Calculate the volume of 0.100 M H2CO3 required to neutralize 25.0 mL of 0.200 M KOH.

1H2CO3 + 2KOH → K2CO3 + 2HOH? L 0.0250 L0.100 M 0.200 M

0.0250 L KOH x 0.200 mol x 1 mole H2CO3 L 2 mole KOH

5. Calculate the volume of 0.100 M H2CO3 required to neutralize 25.0 mL of 0.200 M KOH.

1H2CO3 + 2KOH → K2CO3 + 2HOH? L 0.0250 L0.100 M 0.200 M

0.0250 L KOH x 0.200 mol x 1 mole H2CO3 x 1 L L 2 mole KOH 0.100 mol

5. Calculate the volume of 0.100 M H2CO3 required to neutralize 25.0 mL of 0.200 M KOH.

1H2CO3 + 2KOH → K2CO3 + 2HOH? L 0.0250 L0.100 M 0.200 M

0.0250 L KOH x 0.200 mol x 1 mole H2CO3 x 1 L = 0.0250 L L 2 mole KOH 0.100 mol

6. Label the equation that has a Ka, Kb, Ksp, Kw, or Keq.  

Mg(OH)2(s) ⇌ Mg2+ + 2OH-

 CN- + H2O ⇌ HCN + OH-

NH4+ + H2O ⇌ H3O+ + NH3

  N2O4(g) ⇌ 2NO2(g)

 2H2O ⇌ H3O+ + OH-

6. Label the equation that has a Ka, Kb, Ksp, Kw, or Keq.  

Ksp Mg(OH)2(s) ⇌ Mg2+ + 2OH-

 CN- + H2O ⇌ HCN + OH-

NH4+ + H2O ⇌ H3O+ + NH3

  N2O4(g) ⇌ 2NO2(g)

 2H2O ⇌ H3O+ + OH-

6. Label the equation that has a Ka, Kb, Ksp, Kw, or Keq.  

Ksp Mg(OH)2(s) ⇌ Mg2+ + 2OH-

 Kb CN- + H2O ⇌ HCN + OH-

NH4+ + H2O ⇌ H3O+ + NH3

  N2O4(g) ⇌ 2NO2(g)

 2H2O ⇌ H3O+ + OH-

6. Label the equation that has a Ka, Kb, Ksp, Kw, or Keq.  

Ksp Mg(OH)2(s) ⇌ Mg2+ + 2OH-

 Kb CN- + H2O ⇌ HCN + OH-

Ka NH4+ + H2O ⇌ H3O+ + NH3

  N2O4(g) ⇌ 2NO2(g)

 2H2O ⇌ H3O+ + OH-

6. Label the equation that has a Ka, Kb, Ksp, Kw, or Keq.  

Ksp Mg(OH)2(s) ⇌ Mg2+ + 2OH-

 Kb CN- + H2O ⇌ HCN + OH-

Ka NH4+ + H2O ⇌ H3O+ + NH3

  Keq N2O4(g) ⇌ 2NO2(g)

 2H2O ⇌ H3O+ + OH-

6. Label the equation that has a Ka, Kb, Ksp, Kw, or Keq.  

Ksp Mg(OH)2(s) ⇌ Mg2+ + 2OH-

 Kb CN- + H2O ⇌ HCN + OH-

Ka NH4+ + H2O ⇌ H3O+ + NH3

  Keq N2O4(g) ⇌ 2NO2(g)

 Kw 2H2O ⇌ H3O+ + OH-