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Salts. Since acid + base  salt + H 2 O, when a salt is mixed in water, it dissociates. The cation and anion may be able to absorb H + and OH - ions. This could alter the pH of a solution. To determine if a salt solution is acidic or basic Divide the salt into cation and anion

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slide1

Salts

  • Since acid + base  salt + H2O, when a salt is mixed in water, it dissociates. The cation and anion may be able to absorb H+ and OH- ions. This could alter the pH of a solution.
  • To determine if a salt solution is acidic or basic
  • Divide the salt into cation and anion
  • 2. Recreate the acid and base using H+ and OH-
  • 3. Compare the strengths of the acid and base

strong

strong

neutral

7

strong

weak

acidic

low

weak

strong

basic

high

weak

weak

?????

????

slide2

For weak acids and bases, we look at Ka and Kb of each. The higher value determines if the solution is acidic or basic.

Reminder

Strong acids

HCl HBr HI HNO3 HClO3 HClO4 H2SO4

Strong Bases

Group I hydroxides Heavy group 2 hydroxides

Example – Determine if the following solutions should be acidic or basic.

a. CaCl2 b. NaNO3 c. Na2SO3 d. KC2H3O2 e. NH4Br

neutral

basic

basic

basic

acidic

slide3

Determining the pH of a salt solution

  • Split the salt into its acid and base ions
  • 2. Ignore the strong one (this will not alter the [H+] or [OH-])
  • 3. Treat the cation as a weak acid and the anion as a weak base and write the Ka or Kb expression
  • 4. Use Ka or Kb to determine the concentration of H+ or OH-
  • 5. Determine pH

Examples – Write out the important equilibrium that occurs when we add the following salts to water

a. NaClO b. NH4Cl c. KNO2

  • ClO- + H2O 
  • NH4+ + H2O
  • NO2- + H2O 

HClO + OH-

H3O+ + NH3

HNO2 + OH-

slide4

Examples

If Ka HF = 6.8 x 10-4 and Ka of NH4+ is 5.6 x 10-11, determine the pH of

a. 0.15 M NH4Cl solution b. 0.50 M NaF solution

a. Since NH4Cl will be acidic, treat NH4+ as a weak acid, ignore Cl-.

NH4+ NH3 + H+

0.0 M

0.15 M

0.0 M

-x

+x

+x

0.15

x

x

Ka = [NH3][H+]

[NH4+]

5.6 x 10-11 = x2

0.15

X = 2.90x 10-6

pH = -log 2.90x 10-6

pH =5.54

slide5

For problem B,

NaF  Na+ + F-

ignoretreat as a weak base

F- + H2O  HF + OH-

0.0 M

0.5 M

0.0 M

-x

+x

+x

0.5

x

x

We need Kb for F-, so alter the Ka

Ka x Kb = Kw

6.8 x 10-4 x Kb = 1 x 10-14

Kb = 1.47 x 10-11

Kb = [HF][OH-]

[F-]

1.47 x 10-11 = x2

0.5

X = 2.71x 10-6

pOH = -log 2.71x 10-6

pOH =5.56

pH = 8.44