ACID STRENGTH

Binary acidsGeneral formula HX(aq), where X=Cl, Br, I and F

• Two factors for acid strength trend: • Across a period, electronegativity increases; the acid

strength increases as EN increases • WHY? The more electronegative the atom , the more it draws

electrons away from the hydrogen atom – making it relatively positive• The negative end of a water molecule is then able to easily and strongly

attract to the hydrogen atom and pull it away

•Down a group, bond strength decreases• WHY? A weaker bond means that the hydrogen atom is more easily pulled away from the atom to which it is attached• Ie. HF is a stronger acid than water, but HF is the

weakest of the hydrohalic acids (HCl, HI, HBr, etc.) because the H-F bond is relatively strong!

Oxyacids • Acids that contain oxygen atoms

• Increase in acid strength with increasing number of oxygen atoms • Oxygen is more electronegative than hydrogen, so oxygen atoms

draw electrons away from hydrogen atoms • The more oxygen atoms there are in a molecule, the greater the polarity

making it easier for the water molecule to tear the hydrogen atom away!

• When the central atoms are in the same group and have the same # of oxygen atoms, bond strength increases/acid strength decreases from bottom to top within a group:• eg.) H2SO4(aq) H2SeO4(aq)>

• Oxyacids with the same # of oxygens on the central atom, acid strength increases from left to right in a period: • eg.) H2SO4(aq) H3PO4(aq)>

• For a given central atom, the acid strength of an oxyacid increases with the # of oxygens it holds: • eg.) H2SO4(aq) H2SO3(aq)

• HNO3(aq) HNO2(aq) >>

Strong Acids Moderate Weak Acids

HClO4(aq) H2SO3(aq) HNO2(aq)

HClO3(aq) H3PO4(aq) HF(aq)

HI(aq) CH3COOH(aq)

HBr(aq) H2CO3(aq)

H2SO4(aq) HCN(aq)

HCl(aq)

HNO3(aq)

8.2 WEAK ACIDS & BASES

Weak Acids• H2CO3(aq), HF(aq), CH3COOH(aq), H2CO3(aq), H2S(aq), H3PO3(aq),

H3BO3(aq)

• An acid that partially ionizes in solution but exists primarily in the form of molecules

Percent Ionization• percent ionization (p) =

• For the general weak acid ionization reaction, • HA(aq) H+

(aq) + A-(aq)

• [H+(aq)] = [HA(aq)]

• 1. In a 0.10 mol/L CH3COOH(aq) solution only 1.3% molecules ionize. Calculate the [H+

(aq)]. • CH3COOH(aq) + H+

(aq) • [H+

(aq)] = x 0.10 mol/L• = 1.3 x 10-3 mol/L

• 2. The pH of a 0.050 mol/L methanoic acid solution is 2.78. Calculate the percent ionization of methanoic acid.

• HCO2H(aq) H+(aq) + HCO2

-(aq)

• [HCO2H(aq)] = 0.050 mol/L • pH = 2.78

• [H+(aq)] = 10-pH

• = 10-2.78 = 1.7 x 10-3 mol/L

• p = x 100% • = x 100% • 3.3% • Methanoic acid ionizes 3.3% in a 0.050mol/L solution.

Ionization Constants for Weak Acids• The equilbrium constant for weak acid is known as the

acid ionization constant, Ka.

• HA(aq) + H2O(l) A-(aq) + H3O+

(aq)

• Ka =

• Ie. CH3COOH(aq) + H+(aq)

• Ka =

• 3. Calculate the Ka of hydrofluoric acid, HF(aq) , if a 0.100 mol/L solution at equilibrium at SATP has a percent ionization of 7.8%

• HF(aq) H+(aq) + F-

(aq) • Ka = • Let x be the change in concentration• x = 0.100mol/L x 0.078• x = 0.0078 mol/L

HF(aq) H+(aq) + F-

(aq)

Initial conc. (mol/L) 0.100 0.000 0.000

change in conc. (mol/L) - 0.0078 + 0.0078 + 0.0078

Equilibrium conc. (mol/L) 0.0922 0.0078 0.0078

• Ka = • = • Ka = 6.6 x 10-4

• We can compare the Ka with the values listed in Appendix C9…

• The Ka of hydrofluoric acid at SATP is 6.6 x 10-4

• 4. Butyric acid, HC4H7O2, is used to make compounds employed in artificial flavourings and syrups. A 0.250 mol/L aqueous solution of HC4H7O2 is found to have a pH of 2.72.

• HC4H7O2(aq) H+(aq) + C4H7O2

-(aq)

• .BUT.. x is not unknown, since we have the pH, we can find the [H+(aq)]

• [H+(aq)] = 10-2.72 = 1.9 x 10-3

= x

• Ka = = 1.47 x 10-5

• The Ka of the butyric acid solution is 1.47 x 10-5.

HC4H7O2(aq) H+(aq) + C4H7O2

-(aq)

Initial conc. (mol/L) 0.250 0.000 0.000

change in conc. (mol/L) - X + x + x

Equilibrium conc. (mol/L) 0.250 – x x x

Weak Bases• NH3, N2H4 (hydrazine) • A weak base has a weak attraction for protons

• Base ionization constant, Kb

• B(aq) + H2O(l) HB+(aq) + OH-

(aq) • Kb =

• ie. NH3(aq) + H2O(l) OH-(aq) + NH4

+(aq)

• Kb =

Ionization Constants for Weak Bases

The Relationship between Ka and Kb• For acids and bases whose chemical formulas differ only by a

hydrogen, ie. Conjugate acid-base pairs,

• KaKb = Kw

• 5. What is the value of a base ionization constant, Kb for the acetate ion, C2H3O2

-(aq) at SATP? (*HINT: appendix C9 has Ka

values)• At SATP, Ka of acetic acid is 1.8 x 10-5

• KaKb = Kw • Kb = Kw / Ka = (1.0 x 10-14) / (1.8 x 10-5) • Kb = 5.6 x 10-10

• 6. Calculate the hydrogen ion concentration and the pH of a 0.10mol/L acetic acid solution. Ka for acetic acid is 1.8 x 10-5.

• CH3COOH(aq) + H+(aq)

• Ka =

• Ka = = 1.8 x 10-5 • The unknown can be solved (quadratic equation). Luckily, the

calculation may be simplified

CH3COOH(aq) H+(aq) + CH3COO-

(aq)

Initial conc. (mol/L) 0.10 0.000 0.000

change in conc. (mol/L) - X + x + x

Equilibrium conc. (mol/L) 0.10 – x X x

• The calculation may be simplified by assuming that since Ka is so small, CH3COOH(aq) will ionize very little and the value of x is expected to be very small.

• Will need to use the hundred rule to dtermine whether this assumption is warranted..

• The Hundred Rule: if , a simplifying assumption can be made.

• = 5.6 x 103

• Since 5.6 x 103 > 100,

• The equilibrium equation becomes..• 1.8 x 10-5 • X2 1.8 x 10-6

• x = 1.3 x 10-3

• We must validate our assumption that 0.10 – x = 0.10. Since Ka values are typically known to an accuracy 5%, in general, the approximation will be considered valid if, for the acid HA,

• x = 1.3 x 10-3 , [HA]initial = 0.10 mol/L • = (1.3 x 10-3 mol/L )/ (0.10 mol/L) x 100% = 1.3 % • Since 1.3% < 5%, the assumption we made was valid. Therefore..

(last slide for this answer!!)

• Therefore.. [H+(aq)] = x

• [H+(aq)] = 1.3 x 10-3 mol/L

• pH = -log[H+(aq)]

• = -log(1.3 x 10-3 mol/L)• = 2.89

• 7. What is the pH of a solution that is 0.00250M CH3NH2(aq) ? For methylamine, Kb = 4.2 x 10-4.