Unit 3: Acids, Bases and Buffers

An acid is a substance that donates protons in a reaction, it is a proton donor.

A base is a substance that accepts protons, its is a proton acceptor.

HCl + NH3 --> NH4Cl    or   NH4+  +  Cl-

HCl acts as an acid as it donates a proton (H+) to ammonia.

NH3 acts as a base as it accepts a proton (H+) from HCl.

Acids and bases can only react in pairs. One acid and one base.

HCl + H2O   -->   H3O+   +   Cl-

Here H2O is acting as a base, accepting a proton from HCl which acts as an acid.

H3O+ is called the oxonium ion, or hydronium ion or hydroxonium ion.

Water can also act as an acid. 

H2O   +     NH3   -->  OH-    +    NH4+

Here water donates a proton to NH3 which is acting as a base.

Note: Remember H+ is just a proton. As a hydrogen ion has lost it's one electron and doesn't have a neutron leaving only a proton. H+ can form a bond with any other species that has a lone pair of electrons (this bond will be a co-ordinative bond).

Water is slightly ionised, this is shown by 2 equations:

H2O    ⇌  H+   +   OH-                 and                     2H2O    ⇌    H3O+    +     OH-

As this is an equilibrium, we can write an equilibirum constant equation.

Kc =   [H+] [OH-]   /  [H2O]                                      [something] = concentration of something

[H2O] is a constant so we can modify this equation to:

[H2O] Kc = [H+] [OH-] = Kw

Kw, the water dissociation constant or ionic product of water, is a value used instead of "[H2O] x Kc"

In pure water, at 298K, Kw = 1 x10 to the power of  -14    (1 x 10^(-14) )     

The acidity of a solution depends on [H+] and is measured on the pH scale.

pH = -log[H+]        If you know [H+], you can find pH        (ALWAYS GIVE pH to 2dp!)

The smaller the pH, the greater the [H+] 

As Kw = 1 x10^(-14) = [H+] [OH-] = (1 x10^(-7))   x  (1 x10^(-7))

pH of water at 298K  = -log[H+] = -log(1 x10^(-7)) = 7.00  

H2O ⇌ H+   +   OH-     ΔH = +57.3 kJmol-1      Forwards reaction is endothermic.

Increasing temperature shifts the eqm to the right, so more H+ is produced, increasing [H+] and thus pH. For instance, at 373K the pH of water = 6.00. This does not mean the water is now acidic, but the value for neutrality of water has now changed with the temperature.

REMEMBER! A SOLUTION IS ALWAYS NEUTRAL WHEN [H+] = [OH-]

[H+] can be calculated by doing the opposite of calculating pH.

[H+] = 10 to the power of -pH = 10^(-pH)

For instance if pH = 4.00 then [H+] = 10^(-4)

[H+] can also be found by using the Kw equation.    Kw = [H+] [OH-]

[H+] = Kw / [OH-]

[OH-] can also be found using the Kw equation.

[OH-] = Kw / [OH-]

Strong acid solution:      HCl --> H+   +   Cl-     So in 1.00 moldm-3 HCl, for every mole of HCl there is one mole of H+. So if the concentration of HCl is 1.00 moldm-3 then [H+] once the reaction has occurred = 1.00 moldm-3

pH = -log[H+] = -log(1) = 0.00 which is very acidic from a large [H+]

If it was 0.16 moldm-3 HCl, [H+] = 0.16 moldm-3 and pH = -log(0.16) = 0.80

Alkaline solution: NaOH --> Na +   +   OH-      1.00 moldm-3 NaOH

[OH-] = 1.00 moldm-3                   [H+] = Kw / [OH-] = 1 x10^(-14) / 1 = 1 x10^(-14)

pH = -log[H+] = -log(1 x10^(-14)) = 14.00

If it was 0.1 moldm-3 NaOH, [H+] = 1 x10^(-14) / 0.1 = 1 x10^(-13) pH = -log(1x10^(-13)) = 13.00

Weak acids and bases are not fully dissociated when dissolved in water (molecules split into ions). Weak and strong acids refer to the degree of discociation and so do not change based on concentration. Ethanoic is a weak acid and hydrochloric is a strong acid.

When a weak acid (HA) dissociates to a proton (H+) and conjugate base (A-) we can represent the equilibrium generally as:

HA ⇌ H+ + A-     and thus KC = [H+] [A-] / [HA] = Ka 

Ka is the acid dissociation constant. The greater the value of Ka, the more an equilibrium is to the right. Units for Ka = moldm-3

For weak acids, we can refer to pKa (= -log(Ka) ) which is a measure of the strenght of a weak acid. The smaller the pKa, the stronger the acid.

pH = -log[H+]               pKa = -log(Ka)

[H+] = 10^(-pH)              Ka = 10^(-pKa)                

The equilivalence points is the point at which sufficient alkali has been added to neautralise the acid (or vice-versa). 

High start: strong base, low end: strong acid. Low start: weak base, high end: weak acid.

Low start: strong acid, high end: strong base. High start: weak acid, low end: weak base.

Once worked out the volume of base or acid needed to neautralise an acid or base, you can work out half the volume need to half-neautralise the acid or base. This point is called the half-equilivalence point (or half-end point).

At this point pKa = pH.

At the half end point, as pKa = pH then [H+] = Ka  

Ka = [H+] [A-] / [HA] 

Also at the half-end point, the concentration of the acid, [HA], and conjugate base, [A-], are equal.

As we add a base (OH-) to an acid (HA) the following reaction occurs,

HA          +        OH-         -->         H2O   +         A-

The equilivalance point this reaction, the solution is neautral and so this reaction is complete. But half way through this reaction at the half end point, half of the HA has reacted and half of the A- has formed. As the stoichometry is one to one this means the concentration of acid and conjugate base are EQUAL! THis means [A-] and [HA] cancel out in the Ka equation.

Ka = [H+] [A-] / [HA]

And so Ka = [H+] and thus pKa = pH

21:47

The end point is shown when the volume of alkali or acid added causes the indicator changes colour. Each indicator is set to change when the pH changes within that indicators certain point. e.g. An indicator that changes around pH 7.00 will always change colour when the solution becomes pH 7.00. BUT, all end points are not at pH 7.00 and so different indicators are used to highlight different end points.

Strong acid- strong base: e.g. HCl and NaOH

Either methyl orange or phenolphthalein could be used as the vertical portion of the acid-base titration graph is very long and covers both indicators ranges.

Weak acid- strong base: CH3COOH and NaOH

Phenolphthalein

Strong acid- weak base: e.g.HCl and NH3

Methyl orange

Weak acid- weak base: e.g. CH3COOH and NH3

Here niether indicator would work as the vertical portion of the graph would be too small to notice.

Buffers are designed to keep concentration of hydrogen ions (H+) and hydroxide ions (OH-) almost unchanged.

They are based on equilibrium reactions to remove/add any H+ or OH- added to the system.

Acidic buffers are made from weak acids as the dissociation of a weak acid is an equilibrium reaction.

HA ⇌ H+     +     A-      HA is bigger than H+ and A- (which are equal) as most the acid remains undissociated. So the equilibirum is very far to the left.

If alkali is added: The HA + OH-     --> H2O + A-   which removes the added OH-. THe eqm then shifts more to the left to remove the change in A- concentration.

If acid is added, the extra H+    +   A-   -->  HA  so the eqm shifts to the left. However since A- is in low supply it soon runs out. This is why a soluble salt of HA (the conjugate base) is added to a weak acid as a buffer to increase the supply of A-.

An acidic buffer us made from a mixture of a weak acid and a soluble salt of that acid. It will maintain a pH below 7.00 (acidic). However it's pH will not be constant, adding acids or bases will have an effect on pH, but not a major one.

The function of the weak acid component is to supply HA that can remove any OH-.

HA + OH-   --->    A-     +     H2O

The function of the salt component is to supply A- that can remove any H+.

A-     +     H+    -->   HA

Another way of achieving a mixture of a weak acid and it's salt is by half neautralising the weak acid with an alkali like NaOH.  This buffers pH = pKa. This is also an effiecent buffer.

Basic buffers are made from a weak alkali and it's salt (the reverse of a acidic buffer). Howevr these buffers maintain a pH above 7.00. 

An example is a mixture of aqueous ammonia and ammonium chloride (NH4+Cl-)

Ammonia (aq) removes added H+

NH3 + H+    --->   NH4+

THe ammonium ion, NH4+, removes added OH-

NH4+   +    OH-    -->   NH3    +   H2O

n =     m / Mr   =    n   x V  / 1000    (Could be a salt or an aqueous solution, remember both!)

pH = -log[H+]                     [H+] = 10^(-pH)

pKa = -log(Ka)                     Ka =  10^(pKa)

Kw = [H+] [OH-] for water and all aq solutions          Also   pKw = pH + pOH

Kw = [H+] [H+] for pure water (as [H+] = [OH-] )

Ka = [H+] [A-] / [HA] for weak acid (HA) and buffer calculations

Ka = [H+] [H+] / [HA] for weak acid with nothing added

Ka = [H+] and pKa = pH at the half equalisation/end point.