TOPIC 18 ACIDS AND BASES 18.2 Calculations involving acids and bases

ESSENTIAL IDEA The equilibrium law can be applied to acid-base reactions. Numerical problems can be simplified by making assumptions about the relative concentrations of the species involved. The use of logarithms is also significant here. NATURE OF SCIENCE (1.9) Obtaining evidence for scientific theories – application of the equilibrium law allows strengths of acids and bases to be determined and related to their molecular structure.

INTERNATIONAL-MINDEDNESS Mathematics is a universal language. The mathematical nature of this topic helps chemists speaking different languages to communicate more objectively.

GUIDANCE Know that the value for K w depends upon temperature.

The Ionization of Water The reaction for the ionization of water is: H 2 O (l) ↔ H + (aq) + OH - (aq) So K w = [H + ][OH - ] K w is the ionic product constant of water and does not include water in the expression because water it is a liquid.

In pure water, [H + ] = [OH - ] at 25°C, K w = [H + ][OH - ] = 1.00 x to solve for [H + ], [H + ] = √K w So at 25°C, [H + ]=√(1.00 x )= 1.00x10 -7 pH=-log[H + ]=-log(1.00x10 -7 )= 7.00

Deduce [H + ] and [OH - ] for water at different temperatures given K w values.

K w is Temperature Dependent The dissociation of water is an endothermic reaction because it requires the breaking of bonds. heat + H 2 O (l) ↔ H + (aq) + OH - (aq) An increase in temp shifts equilibrium to the right which increases ion concentrations. A decrease in temp shifts equilibrium to the left which decreases ion concentrations.

Water is always neutral because [H + ] = [OH - ] The pH of water is 7.00 only at 25°C. At higher temperatures, the ion concentrations are higher so pH’s are lower. pH is 6.63 At lower temperatures, the ion concentrations are lower so pH’s are higher. pH is 7.47 The pH’s can fluctuate above and below 7.00, but water is neutral because [H + ] = [OH - ]

To solve for [H + ] of water at any temperature, just take the square root of the given K w. The [OH - ] is then equal to the [H + ].

UNDERSTANDING/KEY IDEA 18.2.A The expression for the dissociation constant of a weak acid is K a and a weak base is K b.

Weak Acids in water HA (aq) + H 2 O (l) ↔ H 3 O + + A - K a = [H 3 O + ][A - ] [HA] or HA (aq) ↔ H + + A - K a = [H + ][A - ] [HA] H + and H 3 O + are used interchangeably.

KaKa K a = acid dissociation or ionization constant The larger the K a, the more the weak acid has dissociated so we can compare strengths of weak acids by their K a ’s.

Weak Bases in water B (aq) + H 2 O (l) ↔ BH + + OH - K b = [BH + ][OH - ] [B] You must ALWAYS include water in the base equation because you have to show that water is the source of H +.

KbKb K b = base dissociation or ionization constant The larger the K b, the more the weak base has dissociated so we can compare strengths of weak bases by their K b ’s.

UNDERSTANDING/KEY IDEA 18.2.B The relationship between K a and pK a is (pK a = -logK a ) and between K b and pK b is (pK b = -logK b ).

pK a and pK b Just as pH is a compact way to measure the H + concentration, pK a and pK b are compact ways to measure K a and K b. K a and K b are often very small numbers and involve negative exponents so the pK a and pK b values are easier to deal with. pK a = -logK a and pK b = -logK b K a = 10 (–pKa) and K b = 10 (-pKb)

The pK a and pK b scale is logarithmic so it compresses a very wide range of K a ’s and K b ’s into a much smaller scale of numbers. This means that if you increase the pK a or pK b by one unit, the K a or K b decreases by 10 times and if you decrease pK a or pK b by one unit, K a or K b increases by 10 times.

pK a and pK b are usually positive and have no units. The relationship between pK a and K a and pK b and K b are inverse. That means as K a increases, pK a decreases. The smaller the K a, the larger the pK a, the weaker the acid. The larger the K b, the smaller the pK b, the stronger the base.

UNDERSTANDING/KEY IDEA 18.2.C For a conjugate acid base pair, K a x K b = K w

More derivations The general acid equation: HA (aq) ↔ H + + A - K a = [H + ][A - ] [HA] The general conj base of the acid equation: A - + H 2 O (l) ↔ HA + OH - K b = [HA][OH - ] [A - ]

When you multiply the 2 equations: K a x K b = [H + ][A - ] x [HA][OH - ] [HA] [A - ] The [HA]’s and [A-]’s cancel, leaving: K a x K b = [H + ][OH - ] = K w

Take the logarithms of both sides: pK a + pK b = pK w We know that at 25°C, K w = 1.00x so pK w = Therefore, pK a + pK b = 14.00

APPLICATION/SKILLS Be able to discuss the relative strengths of acids and bases using values of K a, pK a, K b, and pK b.

HA + H 2 O ↔ H 3 O + + A - HA is the acid with A - its conjugate base. H 2 O is the base with H 3 O + as its conjugate acid. There is a competition for the proton (H+) between the bases. The bases in this example are A - and water. If water is a stronger proton acceptor than A -, then the acid is strong because it dissociates readily and the equilibrium lies to the right. If A- is the stronger proton acceptor, the acid will be weak because most will stay in the HA form and the equilibrium will lie to the left.

A strong acid has a weak conj base. Its conjugate base is weaker than water so water wins the competition for the H + ions. A weak acid has a strong conj base. Its conjugate base is stronger than water. The water is not very successful in pulling the H + ’s from the weak acid.

The higher the K a and K b, the stronger the acid or base because the equil lies farther to the right and it dissociates more. The higher the K a and K b, the lower the pK a and pK b.

APPLICATION/SKILLS Be able to solve problems involving [H + ], [OH - ], pH, pOH, K a, pK a, K b, and pK b.

GUIDANCE When making approximations in weak acid or weak base problems, always state the assumption. Quadratic equations will not be assessed.

RELATIONSHIPS OF pH and pOH pH = -log [H + ] pOH = -log [OH - ] pH + pOH = 14 [H+] = 2 nd log (-pH) [OH-] = 2 nd log (-pOH) Sig Fig Rule for pH – the number of decimal places in the pH figure is equal to the number of sig figs in the original [H + ].

pH’s of Strong Acids and Bases Since the 6 strong acids and the 7 strong bases dissociate or ionize completely, you can calculate the pH directly from the given concentration. 2mol/dm 3 HCl ↔ 2mol/dm 3 H mol/dm 3 Cl- 1mol/dm 3 Ba(OH) 2 ↔ 1mol/dm 3 Ba mol/dm 3 OH -

pH’s of Weak Acids and Bases To calculate the pH of weak acids and bases, you must use equilibrium calculations.

LET’S RE-VISIT ICE TABLES Only molar concentrations are used in ICE tables. ICE stands for “Initial” concentrations, “Change” in concentrations, and “Equilibrium” concentrations.

To solve for pH for weak acids and bases, you will use ICE tables.

Example 1 (Basic weak acid) Calculate the [H + ] and percent dissociation of a 1.0 mol/dm 3 solution of HF (K a = 7.2x10 -4 ). What are the major species? HF and H 2 O Set up your ICE table. HF ↔ H + + F - I C -x +x +x E 1.0-x x x Assume 1.0-x = 1.0 because K a is so small.

Set up your K a expression and solve for x. K a = [H+][F-] = 7.2x10 -4 = x 2 x = [H + ]=.027 [HF] 1.0 Calculate the % dissociation. % dissoc = x =.027 x 100% = 2.7% [conc] ini 1.0

Review example problems on pages 373 and 374 in the HL IB Chemistry textbook.

Citations International Baccalaureate Organization. Chemistry Guide, First assessment Updated Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.