ACID-BASE CHEMISTRY

STRENGTH OF AN ACID OR BASE Strength: The tendency to donate or accept a proton, i.e., how readily does the substance donate or accept a proton? Weak acid has weak proton-donating tendency; a strong acid has a strong proton- donating tendency. Similarly for bases, Cannot define strength in absolute sense. Strength depends on both the acid and base involved in an acid-base reaction. Strength measured relative to some reference, in our case, the solvent water.

THE HYDRONIUM ION The proton does not actually exist in aqueous solution as a bare H + ion. The proton exists as the hydronium ion (H 3 O + ). Consider the acid-base reaction: HCO H 2 O  H 3 O + + CO 3 2- Here water acts as a base, producing the hydronium ion as its conjugate acid. For simplicity, we often just write this reaction as: HCO 3 -  H + + CO 3 2-

STRENGTH MEASURED QUANTITATIVELY BY THE IONIZATION CONSTANT HA 0 + H 2 O  H 3 O + + A - or HA 0  H + + A - The larger K A, the stronger the acid; the smaller K A, the weaker the acid

DEFINITION OF pK A AND pH pK A = - log K A Thus, the larger pK A, the weaker the acid; the smaller pK A, the stronger the acid. Similarly, pH = - log [H + ] pOH = - log [OH - ]

H 2 CO H 2 O  H 3 O + + HCO 3 - NH H 2 O  H 3 O + + NH 3 0 CH 3 COOH 0 + H 2 O  H 3 O + + CH 3 COO - H 2 O + H 2 O  H 3 O + + OH - The stronger an acid, the weaker the conjugate base, and vice versa: The conjugate bases of weak acids are strong, and the conjugate bases of strong acids are weak. CONJUGATE ACIDS-BASES Acids (blue) on left, conjugate bases (green) on right

The strength of an acid is expressed by the value of the equilibrium constant for its dissociation reaction. Consider: HCO 3 -  H + + CO 3 - The dissociation constant for this reaction at 25°C is: This can also be expressed as pK a = 10.3 The larger the pK a, the weaker the acid. Bicarbonate is considered to be a very weak acid. STRENGTH OF ACIDS AND BASES- I = a[H + ] a[CO 3 ] 2- / a[HCO 3 - ]

HCO H 2 O  H 3 O + + CO 3 2- CO 3 2- is a stronger base than H 2 O pK B of H 2 O = 14 pK B of CO 3 2- = 3.7 At high pH: HCO OH -  H 2 O + CO 3 2- (OH) - is a stronger base than CO 3 2- pK B of (OH) - = 0 STRENGTH OF ACIDS AND BASES- II

Explanation: For the reaction, HCO 3- + H 2 O  H 3 O + + CO 3 2- K A = , so pK A = 10.3 K B = K W /K A, so pK B = pK W - pK A = 14 – 10.3 = 3.7 For the reaction, HCO OH -  H 2 O + CO 3 2-, K B for OH - is based on the reaction: OH - + H 2 O  OH - + H 2 O K for the reaction is clearly = 1 Thus, pK B = 0

STRENGTH OF ACIDS AND BASES- IV HNO H 2 O  H 3 O + + NO 3 - HNO 3 is a stronger acid than H 3 O + pKa of HNO 3 is strongly negative pKa of H 3 O + = 0 All acids with pKa smaller than 0, are completely dissociated in water

Most acid-base reactions in aqueous solutions are very fast (almost instantaneous); thermodynamic equilibrium is attained and thermodynamic principles yield correct answers. Kinetics

SELF-IONIZATION OF WATER AND NEUTRAL pH H 2 O  H + + OH - Neutrality is defined by the condition: [H + ] = [OH - ] K w = [H + ] 2 log K w = 2 log [H + ] -log K w = -2 log [H + ] 14 = 2 pH pH neutral = 7 At 25 o C and 1 bar

AMPHOTERIC SUBSTANCE Now consider the acid-base reaction: NH 3 + H 2 O  NH OH - In this case, water acts as an acid, with OH - its conjugate base. Substances that can act as either acids or bases are called amphoteric. Bicarbonate (HCO 3 - ) is also an amphoteric substance: Acid: HCO H 2 O  H 3 O + + CO 3 2- Base: HCO H 3 O +  H 2 O + H 2 CO 3 0

DISSOCIATION CONSTANTS OF SELECTED WEAK ACIDS AT 25°C

EXAMPLES OF ACIDS AND BASES PRESENT IN NATURAL WATERS Most important base: HCO 3 - Other bases: B(OH) 4 -, PO 4 3-, NH 3 0, AsO 4 3-, SO 4 2-, CO 3 2-, etc. Most important acid: CO 2 (aq) or H 2 CO 3 0 Other acids: H 4 SiO 4 0, NH 4 +, B(OH) 3 0, H 2 SO 4 0, CH 3 COOH 0 (acetic), H 2 C 2 O 4 0 (oxalic), etc. Note: B(OH) 3 looks more like an acid written as H 3 BO 3

NUMERICAL EQUILIBRIUM CALCULATIONS Monoprotic acid What are the pH and the concentrations of all aqueous species in a 5 x M solution of aqueous boric acid (B(OH) 3 )? Steps to solution 1) Write down all species likely to be present in solution: H +, OH -, B(OH) 3 0, B(OH) 4 -.

Note on boric acid

2) Write the reactions and find the equilibrium constants relating concentrations of all species: H 2 O  H + + OH - (I) (II) B(OH) H 2 O  B(OH) H +

3) Write down all mass balance relationships: 5 x M =  B = [B(OH) 4 - ] + [B(OH) 3 0 ](III) 4) Write down a single charge-balance (electroneutrality) expressions: [H + ] = [B(OH) 4 - ] + [OH - ](IV) 5) Solve n equations in n unknowns.

EXACT NUMERICAL SOLUTION Eliminate [OH - ] in (I) and (IV) [H + ][OH - ] = K w [OH - ] = K w /[H + ] [H + ] = [B(OH) 4 - ] + K w /[H + ] [H + ] - [B(OH) 4 - ] = K w /[H + ] (V)

Solve (III) for [B(OH) 3 0 ] [B(OH) 3 0 ] =  B - [B(OH) 4 - ] [H + ][B(OH) 4 - ] = K A (  B - [B(OH) 4 - ])(VI) Now solve (V) for [B(OH) 4 - ] - [B(OH) 4 - ] = K w /[H + ] - [H + ] [B(OH) 4 - ] = [H + ] - K w /[H + ] Substitute this into (VI)

[H + ]([H + ] - K w /[H + ]) = K A (  B - [H + ] + K w /[H + ]) [H + ] 2 - K w = K A  B - K A [H + ] + K A K w /[H + ] [H + ] 3 - K w [H + ] = K A  B[H + ] - K A [H + ] 2 + K A K w [H + ] 3 + K A [H + ] 2 - (K A  B + K w )[H + ] - K A K w = 0 [H + ] 3 + (7x )[H + ] 2 - (3.6x )[H + ] - (7x ) = 0 We can solve this by trial and error, computer or graphical methods. From trial and error we obtain [H + ] = 6.1x10 -7 M or pH = 6.21

[OH - ] = K w /[H + ] [OH - ] = / [OH - ] = M [B(OH) 4 - ] = [H + ] - K w /[H + ] [B(OH) 4 - ] = 6.1x x10 -8 [B(OH) 4 - ] = 5.94x10 -7 M [B(OH) 3 0 ] =  B - [B(OH) 4 - ] [B(OH) 3 0 ] = 5x x10 -7 M = 4.99x10 -4 M

APPROXIMATE SOLUTION Look for terms in additive equations that are negligibly small (multiplicative terms, even if very small, cannot be neglected. Because we are dealing with an acid, we can assume that [H + ] >> [OH - ] so that the mass balance becomes: [H + ] = [B(OH) 4 - ] and then [B(OH) 3 0 ] =  B - [H + ]

(ii) [H + ] 2 = K A ·  B-K A [H + ] [H + ] 2 + K A [H + ] - K A ·  B = 0 This is a quadratic equation of the form: ax 2 + bx + c = 0 and can be solved using the quadratic equation

In our case this becomes: Only the positive root has any physical meaning. [H + ] = 5.92 x We could have made this problem even simpler. Because boric as is a quite weak acid (i.e., very K A value, very little of it will be ionized, thus [B(OH) 3 0 ] >> [B(OH) 4 - ]  B  [B(OH) 3 0 ] = 5 x M

[H + ] 2 = 3.5 x [H + ] = 5.92 x M It is wise to check your assumptions by back substituting into original equations. If the error is  5%, the approxi- mation is probably justified because K A values are at least this uncertain!