acids, acids everywhere

Acids, Acids EverywhereAcids and or systems that are controlled by acidity are in almost every aspect of our day to day lives.

From:Our morning coffee…(a base)Digestion of food in our stomachs.To:The production of soap and shampooThe gas we put in our cars

Pretty much everything is affected/created by acids and bases.

Equilibrium and Acid/Base Strength

In our previous work with acids and bases we learned that certain acids are stronger than others.

6 Strong Acids?:

HClHBrHIHNO3H2SO4HClO4

Hydro Chloric AcidHydro Bromic AcidHydro Iodic AcidNitric AcidSulfuric AcidPerchloric Acid

Names?So why are certain acids strong and other are weaker???

Previously we have learned that in order to produce and acid or basic solution chemicals have to react with water to produce?(ions)?

H3O+ Hydronium ACIDSOH- Hydroxide BASES

Now that we have looked at the equilibrium that exists between reactants and Products we can use this knowledge with acids/bases.

H20 H30+, OH-

Even pure water has an equilibrium between the liquid (H2O) and ion(H3O+/OH-) forms.

Because of this equilibrium, even pure water has some H+ and OH- ions in it.

The amount of ions present (%ionization) in the equilibrium of water is very small, something like 2 out of 1,000,000 molecules.

The % ionization (the chance ions are created) is so small that the corresponding Kc is also very small.

Because water is involved in every solution on earth its very important and has a special symbol and it equilibrium constant is very important.

If you write the Kc expression for the ionization of water you notice something familiar.

H2O(l) H3O+(aq) + OH-

(aq)2

[H3O+(aq)][OH-

(aq)]

[(H2O(l)]2 No liquids!

pH & pOH…..againWhenever you deal with pH and pOH, all you need is!..... ???

Perfect Diamond

Shape

Practice

We’ve seen previously that a 1.0mol/L concentration of an acid can have a pH anywhere from -1 to 6.9

So it’s the same concentration why does and its an acid why doesn’t it have the same pH?

***The concentration of the acid has nothing to do with how strong it is. (its pH…how many H3O+

(aq) ions it has).***

If an acids strength is determined by how many H3O+(aq)

then really acid strength is determined by the equilibrium between:

Acid-H(aq) + H2O(l) H3O+(aq)

%?

Calculating Concentration of H3O+(aq) Using % Equilibrium

HCl(aq)

It’s a strong acid so its percent reaction is 100%.

C=0.1 mol/L C=?

100%

Given Information Question

[H3O+(aq)] = %Reaction X [HCl(aq)] = (1) X 0.10 mol/L = 0.10 mol/L

C=0.10 mol/L C=0.10 mol/L

?1:1 Ratio

Just saying…..

Keep that in mind

Hint Hint…

Practice

[H3O+(aq)] = %Reaction X [CH3COOH(aq)]

= (0.013) X 0.10 mol/L

= 0.0013 mol/L

~ 1.3 x10-3 mol/L

% Reaction must be in a decimal!

Re-areanging the % Reaction Formula

As with all formulas, you can also re-arrange the % reaction formula to solve for any of the three variables.

[H3O+(aq)] = %reaction X [Acid-H(aq)]

[H3O+(aq)] = %reaction[Acid-H(aq)]

[Acid-H(aq)] [Acid-H(aq)]

(100) X

But what about compounds that can accept AND donate H+

(aq)?H2CO3(aq) Sodium BiCarbonateH2CO3(aq) + H2O(l)HCO3

-(aq) + H3O+

(aq) HCO3

-(aq) + H2O(l) H2CO3

2-(aq) + OH-

(aq)HCO3

-(aq) + H2O(l) CO3

2-(aq) + H3O+

(aq)

So How do we know what its going to do? Acid or Base?

AcidBase

The Proton Transfer ConceptTheory For Acids and Bases

(Bronsted-Lowry Definition of Acids and Bases)

Johannes Bronstead (1879-1947

Thomas Lowry

(1874-1936

Acids: Donate Protons (H+) and turn H2O into H3O+ ions.

Bases: Accepts Protons turning H2O molecules into OH- ions.

Amphoteric: Is an entity that can act as an acid (donate protons) or a base (accepting protons).

Amphiprotic: Is a entity that can donate/accept ONE SINGLE proton.

Bronsted-Lowry ACIDSProton Donors

Little Extra: The H2O accepts the proton…so by the definition it is a what?

In a chemical equation where an acid donates a proton their MUST ALSO be a BASE to accept it….ALWAYS.

Bronsted-Lowry BASESProton Acceptors

Little Extra: Again, if H20 donates the proton then NH3 must accept it….making the NH3 a what???

And yet again, there are ALWAYS an acid AND a base.

Labeling Bronsted-Lowry Practice

So if an acid donates an H+ and a base accepts an H+

(aq)……does a base turn into an acid after it accepts an H+

(aq )???And vice versa when an acid donates its proton??

Conjugate Acids and BasesWhen an acid reacts in solution it donates its H+

(aq) to the H2O(l) to create H3O+(aq).

H2O(l) + H1+

(aq) H3O+

(aq)

But what happens to the other half of the acid after the H is gone?

HCl(aq) + H20(l) H3O+ + Cl-(aq)

Acid

Conjugate BaseBase

Conjugate Acid

Identifying Conjugate Acids/Bases

Acetic acid (vinegar) is mixed with water and ionizes to produce hydronium ions. But because acetic acid is such a weak acid only a small % of the acid ionizes to produce hydronium ions in the following equation. Label the different chemicals in the equation as; acid, conjugate base, base, conjugate acid.

Acid

Conjugate Base

Base

Conjugate Acid

Acid & Base Conjugates General Form

In general acids contain H’s (at least one) in their chemical formula’s.

When acids react in a chemical reaction they donate the H+

(aq) to either:Water to form H3O+

(aq).Or a base to neutralize OH-

(aq) molecules, which forms……H2O(l).Acid General Form:

Use to be neutral.Lost H+

(aq).Leftover is 0-(+1)= -1

Predicting Acid-Base ReactionsAcid/Base StrengthSo the big question now is;

When you mix a bunch of chemicals together, a lot of chemicals can act as either an acid (accepting) or a base….how do we know which chemicals are going to do what?

Lots of different chemicals can act as a base, and lots of chemicals can act as an acid.

How do we know which chemicals will act like an acid and which will act like a base?

Survival of the fittest…..err strongest.

Predicting Acid-Base ReactionsAcid/Base StrengthSo we know there are six STRONG acids….so

they win.What about the rest that aren’t “Strong”?

Pages 8 and 9 In your data booklet rates the strengths of all acids and bases.

Don’t memorize, just familiarize….

AKA, you will always have the data booklet to help you remember, but! ***Memorizing the most common ones we use will be helpful if you do any chemistry later in life***

STRONG AcIDS

Strong BasesANYTHING WITH OH (hydroxide) IN IT IS A STRONG BASE!

DECREASING STRENGTH

HTGNERTS GNISAERCED

5 Steps To Predicting Acid/Base Reaction Equations1.

Use solubility rules to list all entities present.Sodium Hydroxide (NaOH) is mixed with Vinegar (CH3COOH).

NaOH CH3COOH H2O(l)

Solubility TableIts Ionic

(?) (?)(aq)

Na+(aq) OH-

(aq)

(aq)

Its Molecular

It’s a weak Acid

CH3COOH(aq)H2O(l)

5 Steps To Predicting Acid/Base Reaction Equations2.

Use the acid/base strength table to label each entity as acid (A) or base (B).

Na+(aq) OH-

(aq) CH3COOH(aq)H2O(l)

BA

BA

Acid/Base Table

5 Steps To Predicting Acid/Base Reaction Equations3.

Use the acid/base strength table to label the strongest acid (SA) and strongest base (SB).

Na+(aq) OH-

(aq) CH3COOH(aq)H2O(l)

B

AA

BSB

SA

5 Steps To Predicting Acid/Base Reaction Equations4.

Write a balanced chemical equation with conjugate base (acids ½) and conjugate acid (base ½). Also showing the proton transfer. SA + SB SA(conjugate) + SB(conjugate)

Na+(aq) OH-

(aq) CH3COOH(aq)H2O(l)

A

BSB

SA

CH3COOH(aq) + OH-(aq) CH3COO-

(aq) + H2O(l)

H+

5 Steps To Predicting Acid/Base Reaction Equations5.

Predict the position of the equilibrium….write % of reaction over the reaction arrow.

To Predict the % reaction you use the following generalisation:

CH3COOH(aq) + OH-(aq) CH3COO-

(aq) + H2O(l)

H+Acid/Base Table

5 Steps To Predicting Acid/Base Reaction Equations5.

Predict the position of the equilibrium….write % of reaction over the reaction arrow.

To Predict the % reaction you use the following generalisation:

CH3COOH(aq) + OH-(aq) CH3COO-

(aq) + H2O(l)>50%

DONE!FAIT!

עושה !אניЯ сделал!

A SummaryThe 5 Step Program…Like A.A

First Step is Denial.

Second Step is Anger.

The third step is Bargaining.

The fourth step is Depression.

The fifth step is Acceptance.

These are the 5 steps to overcoming grief.

These have absolutely nothing to do with Chemistry.

Who wrote them down?

A SummaryThe 5 Step Program…Like A.A

Paraphrase it….DONT WRITE IT ALL DOWN WORD FOR WORD!!!

When you talk about acid and base strength you show it just like you did with Kc.

But! Instead of Kc;Ka for AcidKb for BasesThe formula is still the same. PRODCUTS/REACTANTS.

[PRODUCTS][REACTANTS]

Ka and Kb EquationsKa and Kb equations still work the same….

You still set them up the same and use the balancing coefficients as exponents.

You can still put in the information you know from the question and solve for what you don’t know.

Just rearrange it.

Practice!

Well, its an acid so Ka.

Ka=[PRODUCTS] [REACTANTS]

The Mission:Write an equilibriumn expression for hydroflouric acid reacting in water to produce hydronium and HF’s conjugate base.

=[H3O+(aq)][F-

(aq)] [HF(aq)][H20(l)]

Done. Right?

Ka and Kb AND I.C.E Tables.This is just more practice of what we already learned earlier.

Because K a and Kb are the same as Kc, using an I.C.E table is just doing the same as before.

REMEMBER! Whenever your given an equilibrium question, Kc, Ka, Kb, Kwhatever….you need a balanced equation!

Practice!Easy…..(5/10)

Wants Ka…what’s the first thing we need to do?

Ka=[PRODUCTS] [REACTANTS] =[H3O+

(aq)][CH3COO-(aq)]

[CH3COOH(aq)]][H20(l)]

So now just substitute what you know from the question into the equation you just made.

It’s just that easy…….right…..

Practice!

=[H3O+(aq)][CH3COO-

(aq)] [CH3COOH(aq)]

=[H3O+(aq)][CH3COO-

(aq)] [1.00]

Do We know anything else???

[H3O+(aq)]= 10-pH

[H3O+(aq)]= 10-2.38

[H3O+(aq)]= 0.0042 mol/L

????

=[0.0042][CH3COO-(aq)]

[1.00]

What about CH3COO-???

Well since we have a balanced chemical equation we can calculate CH3COO-...(1/1)=[0.0042][0.0042]

[1.00] =0.000017~ 1.7 x 10-5

Practice!Hard-ish (8/10)

What’s the first thing we want to do/make/write down?Balanced Chemical Equation.

Hydrocyanic Acid(aq) + H2O(l) H3O+(aq) + Hydrocyanic Acid –H(aq)

How can we find this chemical formula?

HCN(aq)

How can we find this chemical formula?

CN-(aq)

Practice!Balanced Chemical Equation.

+ H2O(l) H3O+(aq) +HCN(aq) CN-(aq)

Next?Write a Ka expression for the reaction.Ka=[PRODUCTS] [REACTANTS]

=[H3O+(aq)][CN-

(aq)] [HCN(aq)]][H20(l)]=[H3O+

(aq)][CN-(aq)]

[HCN(aq)]

Ka=[H3O+(aq)][CN-

(aq)] [HCN(aq)]

????

6.2 X 10-10=[H3O+(aq)][CN-

(aq)] [HCN(aq)]

????

6.2 X 10-10=[?][CN-(aq)]

[HCN(aq)]????

6.2 X 10-10=[?][?] [HCN(aq)]

????

6.2 X 10-10=[?][?] [0.500]

Well, that’s not a lot to go by.

But because we have more than one unknown, it’s a sign to use an I.C.E table.

Practice!+ H2O(l) H3O+(aq) +HCN(aq) CN-

(aq)

6.2 X 10-10=[?][?] [0.500]

What do we know from the question?Does the question tell us anything about H3O+ or CN-??

What about the change?Are they going to increase or decrease as the reaction occurs?

6.2 X 10-10=[x][x] [0.500-x]

So….

***Because the Ka is so small….6.2 x10-10

=0.00000000062WE ASSUME 0.500-x = 0.500 still

6.2 X 10-10=[x][x] [0.500]

Practice!

6.2 X 10-10=[x][x] [0.500]

Now that we have two unknowns and their BOTH X (because their the same amounts…..because of the 1:1 ratio that we got from the balanced chemical equation.

6.2 X 10-10= X2__ [0.500]

Re-arrange and Solve for X.(500) (500)

0.00000031= X2

X2 = 0.00000000031X = 1.760681686 x 10-5~ 1.76 x 10-5

Practice!~ 1.76 x 10-5

[H3O+(aq)]=1.76 x 10-5

pH = -log[H3O+(aq)]

pH = -log(1.76 x 10-5)pH = 4.754319153

~ 4.75

Water Ionization ConstantKw…Again…

A little while back we learned what Kw is...Anybody remember what it is? The value of it?

So, Kw is the water ionization CONSTANT, so that means its always 1.0 x 10-14.

1.0 x 10-14 might look familiar because we also learned about it with pH and pOH.

Water Ionization ConstantKw…Again…

But now that we know more about equilibrium constants and acids and bases we can change the formula a bit more.

But now that we know that Ka is related to [H+

(aq)]/[H30+(aq)] and Kb is related to we

can make a NEW formula! Kw = Ka X Kb

Like all formulas, this one can be re-arrange to solve for what you don’t know.

This is really useful because if you ever know ka or kb, you can find the other and vice versa.

Write it down!

Ka/Kb Practice

As always, the first step is:Balanced Chemical Equation

So where would we look for find the Kb?

But the data booklet “Relative Strengths of Acids and Bases” table on has Ka’s…So use the water ionization formula.Put in what you know.

Solve for the one you don’tKb in this case.

Kw = Ka X Kb

1.0 x 10-14 = 5.6 x 10-10 X Kb

KbKa

Kb 1.0 x 10-14 5.6 x 10-10

= 1.785714286 x10-5~ 1.8 x 10-5

Big things to remember about titration:

Titrant

Standard Solution

End Point

Equivalence Point

The stuff being dripped out of the burette.

The solution in the beaker that you KNOW the concentration of.

Is the point, the exact drop that causes the indicator in the standard solution to change color.

THEORETICAL point at which the amount of titrant moles = Standard solution moles.

pH Curve’s Unique ShapeBreaking It Down Into Regions

pH curve for Strong Acid (HCl) and Strong Base (NaOH)

1. Buffering Region

2. “The Waterfall” or “Change Region”

3. Buffering Region

• During this time OH and HCl are reacting (canceling each other out) making H2O

• Not a big change in pH

• At this point the last moles of NaOH are being neutralized by added moles of HCl.

• After the last moles of NaOH are gone, a single drop of HCl makes a huge change in pH.

pH Curve’s Unique ShapeBreaking It Down Into Regions

pH curve for Strong Acid (HCl) and Strong Base (NaOH)

3. Buffering Region• All NaOH is gone, all

that’s left is acid.• The pH levels off at

the pH of the pure acid.

• HCl 0.8-1.7• CH3COOH 2.4-3.4• H2SO4 1.1-1.9• The pH range

depends on how concentrated the acid is

% Reaction Slide