…a lesson in resistance

Pretend that you are making a 0.750 M solution of benzoic acid. Another name for benzoic acid is _______________

hydrogen benzoate

Hydrogen benzoate is what kind of acid? _________________An organic acid

Write the acid equilibrium reaction between benzoic acid and the water into which it’s dissolved.

Determine the [H3O+]

…a lesson in resistance

X = [ H O ] =0.0069

3 +

You know the next task….pHind the pH !!!

The pH of this weak acid solution is 2.16

…a lesson in resistance

Name a salt of the conjugate base that could be added to the benzoic acid that would not add to the acidic nature of the solution…. One possible answer would be NaC7H5O2

If you added solid sodium benzoate to the solution such that the concentration of the hydrogen benzoate and the sodium benzoate turned out to 0.750M each, what major species would be present?

…a lesson in resistance

HC7H5O2 ,Na+ ,C7H5O2- ,H2O What would

the concentrations of these species be, that are of any importance, before any reaction occurs?

0.750M

Doesn’t matter, won’t influence things

0.750 M, you can neglect the contribution from the weak acid.

Doesn’t matter, won’t influence things

What reaction contains both of the major species identified?

…a lesson in resistance

Once again, pHind the pH _____

Ka equals the hydronium ion concentration…i.e. [H3O+] = 6.4 x 10-5

4.19

…a lesson in resistance

Find the pH of a solution to which 1.0 mL of 6.0 M HCl was added to 250.0 mL of water.

Hints: 1. HCl is strong acid2. Find the number of moles of HCl present…that’s the same as the H3O+ ion3. Divide the number of moles of H3O+ by the new volume it is in to find [H3O+]

…a lesson in resistance

What is the pH of the water with the 1.00 mL of HCl? ____

Did the water resist a change in pH with the addition of a small amount of acid? ____

What is the pH of pure water? _____7.00

1.62

No !!!

…a lesson in resistance

If you were to add 1.0 mL of 6.0 M HCl to 250 mL of the solution you previously made from the benzoic acid and sodium benzoate, what would happen to the pH?

…a lesson in resistance

Always do the stoichiometry first… 0.0060 moles of HCl will react with 0.0060 moles of benzoate in the following reaction ( 1 to 1 mole ratios):

Ask yourself, what is the HCl most likely to react with that is present as a major species (hint: look for the strongest base around)

_____________

Therefore, 0.0060 moles of benzoic acid is produced as well, as this is a reaction with a strong acid and will go to completion.

The benzoate ion

HCl + C7H5O2- --> HC7H5O2 +

Cl-

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HC7H5O2 + H2O H3O+ + C7H5O2

-

(0.773 M) (0.725 M)

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Once again, phind the pH _____

You have easily found the hydronium ion concentration to be 6.8 x 10-5 .

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Did the solution resist a change in pH with the addition of a small amount of acid? _________________The answer is …a resounding

yes!!!

4.19

4.17What was the pH of the solution after you added the acid? ____

What was the pH of the solution before you added the acid? ____

…a lesson in resistance

“What is it that a solution of a weak acid and a salt of its conjugate base is able to do upon the addition of a small amount of even a strong acid?The answer is …resist a change in its pH… something water could not do!!!!!I guess you know that a solution that can do that is called a “buffer”.

…a lesson in resistance

What is the pH of a buffer solution that is 1.25 molar HNO2 and KNO2 respectively?

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HNO2(aq) + H2O(l) H3O+(aq) + NO2

-

(aq)

Init: 1.25 M ~0 1.25 Mchange: -x +x +xEq: 1.25 M-x x 1.25 M+xKa = [x] [ 1.25 + x ] = 4.0 x 10-4

[ 1.25-x ] [x] = 4.0 x 10-4

pH = 3.40

What if you had 50.0 mL of the buffer and added 2.00 mL of 2.00 M HNO3 to it…what would the pH become as a result?

…a lesson in resistance

Original Question: What if you had 50.0 mL of the buffer and added 2.00 mL of 2.00 M HNO3 to it…what would the pH become as a result?Always do the stoichiometry

first!Moles of HNO3 (a strong acid):

(0.00200L)(2.00 mol/liter)= 0.00400 mol HNO3Moles of HNO2 (the weak acid in the buffer):

(0.0500L)(1.25 mol/liter)= 0.0625 mol HNO2

Moles of NO2- (the [conj] base in the buffer):

(0.0500L)(1.25 mol/liter)= 0.0625 mol NO2-

…a lesson in resistance

Original Question: What if you had 50.0 mL of the buffer and added 2.00 mL of 2.00 M HNO3 to it…what would the pH become as a result?Always do the stoichiometry first! The strong acid will seek the strongest major species base that is present… Therefore it will react with the NO2

-:

This reaction goes to completion because the hydronium really wants to give up its proton!!!

Hence, the nitrite goes down by the number of moles of strong acid added and the nitrous acid goes up by the number of moles of strong acid added

NO2-(aq) + H3O+

(aq) -> H2O (aq) + HNO2

(aq)

…a lesson in resistance

Original Question: What if you had 50.0 mL of the buffer and added 2.00 mL of 2.00 M HNO3 to it…what would the pH become as a result?

NO2-: 0.0625 mol – 0.00400 mol = 0.0585

molHNO2: 0.0625 mol + 0.00400 mol = 0.0665 molEach of these moles now exist in 52.0 mL of solution (the original 50.0 mL and the extra 2.00 from the strong acid addition)Therefore: [NO2

-] = 0.0585 mol/0.052 L = 1.13 M

[HNO2] = 0.0665 mol/0.052 L = 1.28 M

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HNO2(aq) + H2O(l) H3O+(aq) + NO2

-

(aq)

Init: 1.28 M ~0 1.13 Mchange: -x +x +xEq: 1.28 M-x x 1.13 M+xKa = [x] [ 1.13 + x ] = 4.0 x 10-4

[ 1.28-x ] [x] = 4.5 x 10-4

pH = 3.35

Did the addition of the strong acid make the pH of the solution of the weak acid and salt of its conjugate base change much?

Now do the equilibrium problem

NO!

Previous pH = 3.40

…a lesson in resistance

If a buffer can be made from a weak acid and a salt of its conjugate base then what other scenario can you envision?

…a weak base and a salt of it’s conjugate acid. Like… ammonia and ammonium chloride in solution.

…a lesson in resistance

Find the pH of the buffer solution containing 0.50 M NH3 and NH4Cl. Then find the pH of the solution after 1.0 mL of 0.75 M NaOH was added to 100.0 mL of it. Also, find the pH of 100.0 mL of pure water after the same addition (1.0 mL of NaOH)

…a lesson in resistance

NH3(aq) + H2O(l) OH-(aq) + NH4

+

(aq)

Init: 0.50 M ~0 0.50 Mchange: -x +x +xEq: 0.50-x x 0.50 M+xKb = [x] [ 0.50 + x ] = 1.8 x 10-5

[ 0.50-x ] [x] = [OH-]= 1.8 x 10-5 pOH = 4.74

pH = 9.26

Now find the pH of the solution after 1.0 mL of 0.75 M NaOH was added to 100.0 mL of this solution.

First Initial Question: Find the pH of the buffer solution containing 0.50 M NH3 and NH4Cl.

…a lesson in resistance

Always do the stoichiometry first!Moles of NaOH (a strong base):

(0.00100L)(0.75 mol/liter)= 0.00075 mol NaOHMoles of NH3 (the weak base in the buffer):

(0.100L)(0.50 mol/liter)= 0.050 mol NH3

Moles of NH4+ (the [conj] acid in the

buffer):

(0.100L)(0.50 mol/liter)= 0.050 mol NH4

+

Second Question: Find the pH of the solution after 1.0 mL of 0.75 M NaOH was added to 100.0 mL of the 0.50 M buffer of NH3/NH4Cl.

…a lesson in resistance

Always do the stoichiometry first! The strong base will seek the strongest major species acid that is present… Therefore it will react with the NH4

+:

This reaction goes to completion because the hydroxide really wants to take a proton!!!Hence, the ammonium goes down by

the number of moles of strong base added and the ammonia goes up by the number of moles of strong base added

NH4+

(aq) + OH-(aq) -> H2O

(aq) + NH3

(aq)

Original Question: Find the pH of the solution after 1.0 mL of 0.75 M NaOH was added to 100.0 mL of the 0.50 M buffer of NH3/NH4Cl.

…a lesson in resistance

Always do the stoichiometry first! Now you can find the new concentrations!NH4

+: 0.050 mol – 0.00075 mol = 0.049 molNH3: 0.050 mol + 0.00075 mol = 0.051 molEach of these moles now exist in 101.0 mL of solution (the original 100.0 mL and the extra 1.00 from the strong base addition)Therefore: [NH4

+] = 0.049 mol/0.101 L = 0.49 M

[NH3] = 0.051 mol/0.101 L = 0.50 M

Original Question: Find the pH of the solution after 1.0 mL of 0.75 M NaOH was added to 100.0 mL of the 0.50 M buffer of NH3/NH4Cl.)

…a lesson in resistanceInit: 0.50 M ~0 0.49

Mchange: -x +x +xEq: 0.50 M-x x 0.49 M+xKb= [x] [ 0.49 + x ] = 1.8 x 10-5

[ 0.50-x ] [x] = [OH-]= 1.8 x 10-5

pOH = 4.74

Now do the equilibrium problem

NH3(aq) + H2O(l) OH-(aq) + NH4

+

(aq)

pH = 9.26

No appreciable change!

…a lesson in resistance

Moles of NaOH (a strong base):

(0.0010L)(0.75 mol/liter)= 0.00075 mol OH-0.00075 mol OH- / 0.101 L = [OH-]

0.0074 M = [OH-]

Final Question Also, find the pH of 100.0 mL of pure water after the same addition (1.0 mL of NaOH)

pOH= 2.13

pH= 11.87 a significant change from pH = 7 in

pure water!