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ACIDS AND BASES. Dissociation Constants. Write the equilibrium expression ( K a or K b ) from a balanced chemical equation. Use K a or K b to solve problems for pH, percent dissociation and concentration. Additional KEY Terms.

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ACIDS AND BASES

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Acids and bases

ACIDS AND BASES

Dissociation Constants


Acids and bases

  • Write the equilibrium expression (Kaor Kb) from a balanced chemical equation.

  • UseKaor Kb to solve problems for pH, percent dissociation and concentration.

Additional KEY Terms


Acids and bases

HA(aq) H+(aq) + A-(aq)

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Strong Acid

Weak Acid

Ka - acid dissociation constant

Larger Ka : stronger acid : more product : more H+


Acids and bases

BOH(aq) B+(aq) + OH-(aq)

B(aq) + H2O(l) BH+(aq) + OH-(aq)

Strong Base

Weak Base

Kb - base dissociation constant

Larger Kb : stronger base : more product : more OH-


Acids and bases

CH3COOH(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq)

Initially a 0.10 M solution of acetic acid, it reaches equilibrium with a [H3O+] = 1.3 x 10-3 M. What is the acid dissociation constant, Ka?

I0.10 0 0

C-1.3 x 10-3+1.3 x 10-3+1.3 x 10-3

Ka = 1.7 x 10-5

Ignore the units for K.

1.3 x 10-3

E0. 0987 1.3 x 10-3

Type III – all initial and one equilibrium concentration


Acids and bases

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

Ka = [H3O][A-]

[HA]

HA is a weak acid with a Ka of 7.3 x 10-8. What are the equilibrium concentrations if the initial [HA] is 0.50 M?

[I]0.50 0 0

[C]-x+x+x

[E]0.5-x +x +x

Type IV– all initial and NO equilibrium


Acids and bases

7.3 x 10-8 = [x][x]

0.50

*Ka is small - assume that x is

negligible compared to 0.50

- x

(7.3 x 10-8)(0.50) = x2

3.65 x 10-8 = x2

[H3O+] = [A-] = 1.9 x 10-4 M

1.9 x 10-4 = x

[HA] = 0.50 - x

= 0.50 - 1.9 x 10-4

= 0.49981 M

*Kais small – OK to ignore it

0.50 M


Acids and bases

Calculate the pH of a 0.10 mol/L hydrogen sulfide solution. (Ka=1.0 x 10-7)

H2S (aq) + H2O (l)

Ka = [H3O+][HS-]

[H2S]

  • H3O+(aq) + HS-(aq)

[I] 0.10 00

[C] -x+x+x

[E] 0.10 - xxx


Acids and bases

1.0 x 10-7 = [x][x]

0.10

*Ka is small - xis negligible

- x

(1.0 x 10-7)(0.10) = x2

1.0 x 10 -8 = x2

[H3O+] = [HS-] = 1.0 x 10-4 M

1.0 x 10 -4 = x

pH = - log [H3O+] = - log(1.0 x 10-4)

pH = 4.00


Acids and bases

  • Each acid/base has K associated with it

  • polyprotic acids lose their hydrogenone at a time - each ionization reaction has separate Ka

H2SO4(aq) H+(aq) + HSO4-(aq)

HSO4- (aq) H+(aq) + SO4-2(aq)

Sulfuric acid H2SO4

Ka1

Ka2


Acids and bases

Percent Dissociation

  • Ka / Kb represent the degree of dissociation

  • (how much product has formed)

  • Another way to describe dissociation is by percent dissociation


Acids and bases

CH2O2H (aq) + H2O(l) H3O+(aq) + CH2O2¯(aq)

Calculate the percent dissociation of a solution of formic acid (CH2OOH) if the hydronium ion concentration is .

0.100 M

4.21 x 10-3 M


Acids and bases

Calculate the Kbof hydrogen phosphate ion (HPO42-) if 0.25 M solution of hydrogen phosphate dissociates 0.080%.

HPO42- + H2O H2PO4- + OH-

Use the %diss formula to find [OH-]


Acids and bases

HPO42- + H2O H2PO4- + OH-

Kb = [H2PO4-][OH-]

Kb= [2.0 x 10-4][2.0 x 10-4]

[HPO42-]

0.25

[I]0.25 0 0

[C]-x+x +x

[E]0.25 +x +x

[OH-] = [H2PO4-] = 2.0 x 10-4 M

Kb= 1.6 x 10-7


Acids and bases

  • The smaller the Ka or Kb, the weaker the acid / base

  • percent dissociation describes the amountof

  • acid/base dissociated


Acids and bases

  • CAN YOU / HAVE YOU?

  • Writethe equilibrium expression (Kaor Kb) from a balanced chemical equation.

  • UseKaor Kb to solve problems for pH, percent dissociation and concentration.

Additional KEY Terms


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