PredictinPredicting the pH g the pH of salt of salt

solutionssolutions

Hydrolysis of ionsHydrolysis of ionsHydrolysis refers to a reaction with water (e.g.

splitting water into H+ and OH–)When salts are added to water, pH can changeE.g. when Na3PO4 is added to water, ions form

Na3PO4(aq) 3Na+(aq) + PO43–(aq)

These ions may react with H2O, affecting the pH

PO43–(aq) + H+(aq) HPO4

2–(aq)

Na+(aq) + OH–(aq) NaOH (aq)If the anion (-ve) reacts to remove lots of H+ but

the cation (+ve) removes very little OH–, then H+ will decrease and the solution will be basic.

The degree of hydrolysisThe degree of hydrolysis PO4

3–(aq) + H+(aq) HPO42–(aq)

Na+ + OH– NaOH• The problem with writing equilibria this way is

we do not know the strength of the reactions• However, if we reverse the reaction we can

look up Ka and Kb values (pg. 608, 615): HPO4

2– PO43– + H+

NaOH Na+ + OH–

Ka= 4.5 x 10–13

Kb= 55• Small Ka: few products; adding PO4

3– = shift left• Large Kb: mostly products; Na+ has little affect• Thus, adding Na3PO4 will cause more H+ to be

removed, resulting in a basic solution

Accuracy of predictionsAccuracy of predictionsTheoretically, using Ka and Kb values you could

predict the exact pH resulting from a certain salt being added to distilled water.

However, you only need to be able to predict if a solution will be acidic, basic, or neutral.

Note: you can’t judge the pH change solely on the difference between Ka and Kb. Other factors are involved (e.g. the formula of the compound and its molar mass both affect [ ])

Note: hydrolysis refers to reactions with water. Several variations for writing equilibriums exist. However, focusing on how the H+/OH– balance of water is affected is easiest.

Steps in determining pH Steps in determining pH 1. Write the ions that form: e.g. NH4CN

NH4+ + CN –

2. Determine the reaction ions have with water:NH4

++ OH– NH3 + H2O, NH3 + H2O NH4

+ + OH–

CN – + H+ HCN, HCN CN– + H+

3. Look up the Ka of the conjugate acid and the Kb of the conjugate base:

4. Determine if more H+ or OH– is removed:More H+ is removed, therefore BASIC

[NH4+][OH–]

[NH3]= 1.8 x 10–5

Kb =[CN –][H+] [HCN]

= 6.2 x 10–10

Ka =

Buffers - labBuffers - labRead 15.6 (621-623) up to and including special

topic 15.2 (carbonate buffer)Calibrate pH meter, get a plastic bottle with

distilled H2O to rinse your pH meter btw tests

You will use 4 solutions (20 mL of each): distilled water, water + NaC2H3O2 (5 scoops), 0.2 M HC2H3O2, 0.2 M HC2H3O2 + NaC2H3O2

For each, record the initial pH and the pH upon addition of 5, 10, and 15 drops of 1 M HCl

Remake the 4 solutionsFor each, record the initial pH and the pH upon

addition of 5, 10, and 15 drops of 1 M NaOH

HCl H2O NaC2H3O2 HC2H3O2

NaC2H3O2 + HC2H3O2

0

5

10

15NaOH

H2O NaC2H3O2 HC2H3O2

NaC2H3O2 + HC2H3O2

0

5

10

15

HCl H2O NaC2H3O2 HC2H3O2

NaC2H3O2 + HC2H3O2

0 6.9 8.0 2.8 5.1

5 1.9 6.3 2.6 5.0

10 1.7 5.6 2.2 5.0

15 1.6 5.7 2.0 4.9NaOH

H2O NaC2H3O2 HC2H3O2

NaC2H3O2 + HC2H3O2

0 6.9 7.7 2.6 5.3

5 11.1 10.3 2.7 5.2

10 11.3 10.5 3.3 5.2

15 11.4 10.8 3.4 5.3

Buffers - summaryBuffers - summarySolutions with buffers resist changes in pH, when

small amounts of acid or base are addedBuffers are important in blood, cells, resisting the

effects of acid rain on lake ecosystems.A buffer is created when a weak acid is mixed

with a salt that contains the identical ion.Two equilibria contribute to the consistent [H+]

HA H+ + A–

NaA+ Na+ A–

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