Question

1 a) Write the aqueous acid dissociation reactions for an acid and base dissociation reaction for a base according to the Bronsted-Lowry definition

b) Determine conjugate bases of acids and acids of bases

c) Write the equilibrium expression for an acid or a base aqueous dissociation

d) Evaluate strength of an acid or base based on its Ka or Kb or pKa or pKb.

e) Apply Kw at 25oC and at different temperatures.

f) Solve for the pH of strong acids or bases.

g) Solve for the pH, equilibrium concentrations, and/or % dissociation given acid or base concentration of either a weak acid or base.

Answer 1

Bronsted-Loery Acid-base theory :- identifies strong and weak acids and bases based on whether the species accept or donate proton (H⁺), according to the theory , an acid and base reacts with each other, causing the acid to form its conjugate base and the base to form conjugate acids by exchanging a proton.

Aqueous acid and base dissociation :-

Conjugate Acid-Base Pair Bronsted-Lowry Conjugate Acid Base H B HA B A Bronsted-Lowry Conjugate Base Acid Conjugate Acid-Base Pair 1L

Equilibrium expression for dissociation of acid and base :-

1. For acid

H30 ][A ] HA Ka K[H2O] =

2. For base

[ВH"|ОН ] К, — К|Н-0] — [B]

There are related scales in chemistry used to measure how acidic or basic a solution is and the strength of acids and bases. Although the pH scale is most familiar , pKa, Ka, pKb and Kb are common calculations that offer insight into acid-base reactions. Ka,pKaKb and pKb are more helpful for predicting whether a species will donate or accept protons at specific pH value. They describe the degree of ionization of acid or base strength because adding water to solution will not change the equilibrium constant.

Ka is the acid dissociation constant, pKa is the negative log of Ka

HA H20 s A + H30* .HB H20 s B OH In the formulas, A stands for acid and B for base Ka [H+][A-]/ [HA] РКа 3- 1og Ka at half the equivalence point, pH = pКa --1og Ka

Kb 3D [B+][ОН-J/[ВОН] pKb -log Kb A large Kb value indicates the high level of dissociation of a strong base. A lower pKb value indicates a stronger base. pKa and pKb are related by the simple relation: РКa + pkb %3D 14

Summary of effect of temperature on K W pH of water is only 7 at 298K (25°C) Note that at temperatures above and below this, despite changes in pH, water is still a neutral substance as its H*(aq) concentration is equal to its OH (aq) concentration. (Neither acidic or basic!) Temperature should always be stated alongside pH measurements as Kw is temperature dependent! [H] in pure water pH of pure water (-log1o [H) Temperature/°C KW K 1.5 x 10-15 0.39 x 10-7 7.47 3.0 x 10-15 10 0.55 x 10-7 7.27 0.82 x 10-7 20 6.8 x 10-15 7.08 1.00 x 10-7 25 1.0 x 10-14 7.00 1.22 x 10-7 6.92 30 1.5 x 10-14 1.73 x 10-7 40 3.0 x 10-14 6.77 50 5.5 x 10-14 2.35 x 10-7 6.63

pH for strong acid - strong base

Calculate pH of solution when 12cm3 of 0.1 mol dm3 HCl reacts with 5cm3 0.15 mol dm3 NaOH HСI NaCl NaOH H20 12/1000 x 0.1 5/1000 x 0.15 7.5 x 104 reacting moles reacting moles therefore, all NaOH is neutralised, and there is excess HC excess moles HCl= 1.2 x 103 7.5 x 10-4 = 4.5 x 104 1.2 x 103 [HCI] 4.5 x 10-4 17/1000 = 0.02647 . [H] 0.02647 pH -log [H*]= 1.58

What is the pH of a 5.0 x 102mol/L solution of barium hydroxide, Ba(OH)20q? Ba(OH)2 is a strong base Ва(ОН)— Ва* [OH 2 x5.0x 102 mol/L pOHlog1o [OH- =-log10[1.0 x 10-11 pH 3 14 - рОН — 13 20H- (aq) (aq) = =

Key Takeaways: pH of a Weak Acid The pH equation is still the same (pH= log[H*]), but you need to use the acid dissociation constant (K) to find [H*]. There are two main methods of solving for hydrogen ion concentration.

pH calculation questions pH of weak base K-5.62x100. Ammonia is a weak base, so the most convenient approach is to calculate pOH using Kp, and then to convert it to pH From the Brønsted-Lowry theory we know that KaxKb Kw that allows us to calcu late Kb=Kw/Ka=10-14/5.62x10-10=1.78x10-5

A weak acid, HA, has a pKa of 4.756. If the solution pH is 3.85, what percentage of the acid is dissociated? First, convert pKa to Ka and pH to [H+]: Ka=10-4.756= 1.754 x 10-5 [H+] 103.85 = 1.413 x 104 Now use the equation Ka = ([H+] [A"]) [HA], with [H*] [A] [(1.413 x 104 M) (1.413 x 1.754 x 105 104 M)] [HA] [HA] 0.0011375 M Percent dissociation is therefore given by 1.413 x 104 0.00 1 1375 = 0.1242 = 12.42%