Water

Water is the major is the major component of living organisms; about 75-90% of a cell consists of water. Unusual properties of water:

• Boiling point = 100 ^oC. This is very high, compared to -161 ^oC for methane, which has a similar molecular weight as water.
• Density of water is highest at 4^oC. This is unusual since most substances are more dense in their solid state. The implication is of course that ice floates. Think about the concequences if ice would be heavier than liquid water.
• Water has a high heat capacity , i.e. it will take a relatively large amount of energy to increase the temperature of water. As a results large bodies of water will experience far less severe temperature changes than the surrounding atmosphere.
• High surface tension
• Excellent solvent properties.

pH

Because of the strong electronegativity of oxygen and H-bond formation the H-atom in water carries a partially positive charge. By chance the bonding pair in some water molecules may be shifted totally to the oxygen causing the ionzation of water, i.e. the formation of an OH^- and H⁺ ion. However, the likelyhood for this to happen is very small, to be exact: 1 mol of H₂O contains 10^-7 mol of OH^- and 10^-7 mol of H⁺. Water is said to dissociate and the degree of dissociation is a constant. It can be expressed as the dissociation constant (K_diss) .

K_diss = [H⁺] x [OH^-] / [H₂O]

Since only 10^-7 mol of water (out of one mol water) are dissociated, the concentration of undissociated water [H₂ O] does not change significantly and can be taken into the dissociation constant K_diss of water. The new constant K_W is called the ion product of water and has a value of 10^-14.

K_diss x [H₂O] = K _W = [H⁺] x [OH^- ]

Using the logarithm of the equation and multiplying by (-1) we obtain

-log K_W = -log [H⁺] - log [OH^- ]

Using a convention to express : -log as p, the above equation can be rewritten as:

pK_W = pH + pOH or
14 = pH + pOH

Thus   pH = -log [H⁺] and pOH = -log [OH^- ].

In case of water pH = pOH = 7 , i.e. water is neither acidic nor basic, it is neutral.

Acids and Bases

We can define acids as proton donors. For example the H-Cl is such a proton donor. If H-Cl is dissolved in water it dissociates completely to:

[H⁺] + [Cl^- ]

This dissociation is 100%, i.e. no undissociated H-Cl is left. Completely dissociated acids are called strong acids . If you have a 0.1 molar HCl then there are 0.1 mol of Cl^- and 0.1 mol of H⁺, or the

pH = -log [H⁺] = -log 10^-1 = 1

Weak acids are dissociated only to a small degree. Acetic acid dissociates to:

CH₃COOH → H⁺ + CH₃COO^-

K_CH3COOH = [H⁺] x [CH₃COO^- ] / [CH₃COOH]

The dissociation constant (K_CH3COOH) is a constant and its value is:

K_CH3COOH = 1.74 x 10^-5 mol / L

Now we can calculate the pH of 0.1 mol acetic acid:

                       CH3COOH   →  H+  +     CH3COO-

conc. before diss.:   0.1 mol       0 mol       0 mol 

conc. after diss.:    0.1-n mol     n mol       n mol 

K_CH3COOH = 1.74 x 10^-5 = n x n / 0.1 - n = n² / 0.1 - n

n is a very small number (based on the dissociation constant), thus 0.1-n = approx. 0.1

K_CH3COOH = 1.74 x 10^-5 = n² / 0.1

n² = 1.74 x 10^-6
n = 1.32 x 10^-3 Since n = [H⁺] we can calculate the pH

pH = -log [H⁺] = -log1.32 x 10^-3 = 2.88

We can define bases as proton acceptors. For example the Na-OH is such a proton acceptor. If Na-OH is dissolved in water it dissociates completely to:

Na-OH dissociates to [OH^- ] + [Na⁺]

This dissociation is 100%, i.e. no undissociated Na-OH is left. Completely dissociated bases are called strong bases. If you have a 0.1 molar NaOH then there are 0.1 mol of OH^- and 0.1 mol of Na⁺, or the

pOH = -log [OH^- ] = -log 10^-1 = 1

we can calculate the pH from pK_W = 14 = pH + pOH or
pH = 14 - pOH = 14 - 1 = 13

Weak bases produce only very small amounts of OH^- when dissolved in water. The pH of a 0.1 molar ammonia solution can be calculated similar as for acetic acid in the above example:

                      NH3  +  H2O  → [NH4]+  +   OH-

conc. before diss.:   0.1 mol          0 mol      0 mol 

conc. after diss.:    0.1-n mol        n mol      n mol 

Since 0.1 - n = approx. 0.1 and n = [OH^-] we can calculate pOH = 2.87 or pH = 11.2

If the dissociation constant K_[NH4]⁺ for the dissociation of [NH₄]⁺ → NH₃ + H⁺ is given instead of K_NH3 one can calculate K_NH3 = K_W / K_[NH4]⁺

Buffers

Buffers are solutions of weak acids (or bases) and their salts, which can absorb H⁺ or OH^- whithout major changes in the pH of the buffer solution. A constant pH is very important in cellular environments. For example intracellular fluids regulate their pH using a phosphate buffer, whereas the pH of blood is controlled by a carbonate buffer. For example a solution containing :

CH₃COOH and CH₃COO^- Na ⁺ can buffer H⁺ or OH^-

CH₃COO^- + H⁺ → CH₃COOH

CH₃COOH + OH^- → CH₃COO^- + H₂O

The pH of a buffer can be calculated using the Henderson Hasselbach equation, which is simply the equation for the dissociation constant, written in logarithmical form:

K_CH3COOH = [H⁺] x [CH₃COO^- ] / [CH₃COOH]

log K_CH3COOH = log [H⁺] x [CH₃COO^- ] / [CH₃COOH]

log K_CH3COOH = log [H⁺] + log ( [CH₃COO^- ] / [CH₃COOH] )

- log [H⁺] = - log K_CH3COOH + log ( [CH₃COO^- ] / [CH₃COOH] )

pH = pK + log [CH₃COO^- ] / [CH₃COOH] )
or
pH = pK - log ( [CH₃COOH] /[CH₃COO^- ]

pH = pK - log [acid] / [base]    Henderson Hasselbach equation

For example the pH of a solution containing 0.1 mol/L CH₃COOH (acetic acid) and 0.15 mol/L CH₃COO^- Na⁺ (sodium acetate) can be calculated. The dissociation constant for acetic acid is :

K_CH3COOH = 1.74 x 10^-5

pH = pK - log [CH₃COOH] / [ CH₃COO^-] = 4.76 - log 0.1 / 0.15 = 4.95

If a solution contains equimolar amounts of acid and base, then its pH = pK (log 1 = 0). At such a ratio a solution has its best buffering capacity for H⁺ or OH^-. For example the phosphate buffer (pK = 7.2) in the intracellular fluids has the highest buffering capacity at pH = 7.2, which is very close to the physiological pH of 7.0

[H₂PO₄]^- (acid) + OH^- → [HPO₄]^-- + H₂O

[HPO₄]^-- (base) + H⁺ → [H₂PO₄]^-

Problems

1. Rank the following intermolecular forces according to their strength (begin with the strongest interaction): ionic bonds, H-bonds, covalent bonds, hydrophobic interactions, polar interactions
2. What is the pOH of water ?
3. hat is the pH of a 0.01 M strong acid ?
4. What is the pH of a 0.01 M NaOH?
5. Calculate the pH of a 0.01 M acetic acid
6. At what pH does acetic acid have its optimum buffering capacity
7. At what pH does carbonic acid have its optimum buffering capacity
   (K for H₂CO₃ ^___> (HCO₃)^- + H⁺ = 1.7 x 10^-4

Answer to Problems

1. covalent bonds, ionic bonds, H-bonds, polar interactions, hydrophobic interactions
2. [OH^- ] of H₂O = 10^-7 mol/L pOH = -log [OH^- ]= -log10^-7 = 7
3. a strong acid is 100% dissociated. If the acid concentration = 0.01 M than the [H⁺] conc. = 0.01mol/L
   pH = -log [H⁺] = -log10^-2 = 2
4. a strong base is 100% dissociated. [OH^- ] conc. = 0.01 mol/L
   pOH = -log [OH^- ] = -log10^-2 = 2 , pH = 14 - pOH = 14 - 2 = 12
5. K_CH3COOH = 1.74 x 10^-5 = [H⁺] x [CH₃COO^- ] / [CH₃COOH] = n x n / 0.01 - n     n = [H⁺] = 3.38
6. At a pH where the conc. of CH₃COOH (acid) = the conc. of CH₃COO^- (base)     pH = pK - log [acid] / [base], if [acid] = [base], then -log1 = 0 and pH = pK (pK acetic acid = 4.74)
7. at pH = pK, K=1.7x10^-4, pK = -log1.7x10^-4 = 3.77