Bronsted-Lowry acids and bases

The difference between dissociation and ionisation Dissociation refers to a reaction where a molecule or substance breaks apart into smaller units. The units are not necessarily ions, although this is often the case. Ionization generally refers to a reaction which forms ions from an uncharged species.

Defining acids and bases In chemistry, the Brønsted-Lowry theory is an acid-base theory, proposed independently by Johannes Nicolau Brønsted and Thomas Martin Lowry in 1923

Bronsted-Lowry definition of acid A substance behaves as an acid when it: donates a proton (H+) to a base. Acids are proton donors. When acids react with water, hydronium (H3O+) ions are produced. H+ ions cannot exist by themselves HCl + H2O  H3O+ + Cl- acid base H+ is attracted to the negative end of H2O to become H3O+

Bronsted-Lowry definition of a base A substance behaves as a base when it: accepts a proton from an acid (Bases are proton acceptors. When bases react with water, hydroxide (OH−) ions are produced.

Acids and bases HCl is an acid because it donates H+ NH3 accepts H+ and therefore is the base (NH4+ and Cl– then form an ionic compound) Lewis acid: electron pair acceptor Lewis base: electron pair donor H N Cl + -

Acid/base conjugate pairs Conjugate means joined together When an acid and a base react together a conjugate acid and base are formed HNO3 + H2O  H3O+ + NO3- acid base acid 2 base 2 The conjugate pairs are (HNO3 /NO3-) (H2O / H3O+) They differ by a H+

Acid/base conjugate pairs HCN(l) + H2O  CN–(aq) + H3O+(aq) HCN is acid, H2O is base HCN(l) + H2O  CN–(aq) + H3O+(aq) H3O+ is acid, CN– is base A conjugate acid-base pair are two substances that differ from each other by just one proton (H+) HCN and CN– and H2O and H3O+ are conjugate acid-base pairs

Questions Pg 245 Q 1,2,3

Hydrolysis Hydrolysis is a chemical reaction during which an anion reacts with water to produce OH– or a cation reacts with water to produce H3O+ H2O + Cl-  OH- + HCl H2O + H+  H3O+

Acid and base strength Acids and bases have different strengths Some acids donate protons more readily than others The strength of an acid is its ability to donate hydrogen ions to a base. The strength of a base is its ability to accept hydrogen ions from an acid. Strength is different from concentration (pg 248 Figure 14.9)

Strong acids and bases A strong acid will completely ionise in solution (producing many ions) A strong base will accept protons (H+) easily

Weak acids and bases A weak acid does not ionise to any great extent and so contains a larger number of molecules compared with the number of ions produced in solution. Completely ionised Partially ionised

Reversible reaction Reversible arrows in an equation show that the products on the right can react together and produce the left hand side. A chemical equation without a double arrow isn't reversible and can only go in one direction.

Polyprotic acids Polyprotic acids are acids capable of donating more than one proton (H+). Monoprotic HCl 2. Diprotic H2SO4 3. Triprotic H3PO4

Amphiprotic substances Some substances act as acids and bases They can donate or accept protons and are called amphi (meaning both) protic (hydrogen ions) Example water, ammonia and amino acids

Write the formulae of the conjugate bases of the following acids: H2SO4 H2S HS- NH4+ Write the formulae of the conjugate acids of the following bases: OH- HCO3- H2O CN-

pH and pOH pH stands for the potential of hydrogen or concentration of H+ pOH is a measure of the hydroxide ion concentration The acidity of a solution is a measure of the concentration of H+ In a neutral solution there is the same concentration of H+ or H3O+ and OH- Basic solutions have a lower concentration of H3O+ than OH- Acidic solutions have a greater concentration of H3O+ than OH-

Ionic product [H3O+ ] [H+ ] represents the concentration of H3O+ or H + [OH-] represents the concentration of OH- Experiments show that all aqueous solutions contain H + and OH- that the product of their molar concentrations is 10-14M2 at 25C The ionic product is: [H3O+ ]x[OH-] = In pure water [H3O+ ]=[OH-] 10-7M = 10-7M

At 25C a solution is: Acidic if [H3O+ ] > 10-7M [OH-] < 10-7M Neutral if [H3O+ ][OH-] Basic if [H3O+ ] < 10-7M [OH-] > 10-7M

Calculating pH and pOH pH is calculated using the following formula: pH = -log [H+] pOH is calculated using the following formula: pOH = -log [OH-] pH + pOH = 14.0 [H+] = 10 -pH [OH-] = 10 -pOH

pH calculations for weak acids Can we use the pH calculation for weak acids? No Why? Because weak acids have not fully ionized so we do not know the H+ concentration You have to wait till year 12 and you do the equilibrium constant

Calculating pH and pOH Find the pH of a solution of sodium hydroxide that has a pOH of 2 pH = 14 – pOH pH = 14 - 2 = 12 Find the pOH of a solution of hydrochloric acid that has a pH of 3.4 pOH = 14 – pH pOH = 14 - 3.4 = 10.6

Questions Find the pH of 25.0 mL of a 0.045 M (mol/L) solution of HCl. What is the pOH? Note that HCl is a strong monoprotic acid which means that... [HCl] = [H+]= 0.045 M pH = -log [H+] pH = -log 0.045 pH = 1.35 The pOH is given by pH + pOH = 14 We substitute in the pH of 1.35 and get: 1.35 + pOH = 14 So, pOH = 12.65

Questions a) Find the pOH of 0.000685 M solution of NaOH. b) What is the pH of the solution? a)Note that NaOH is strong and monobasic which means that... [NaOH] = [OH-] = 0.000685 M hence, pOH = -log [OH-] pOH = -log 0.000685 pOH = 3.164 b) pOH is 3.164. The pH can be found by using pH + pOH = 14. Substituting in gives us pH + 3.164 = 14 So, pH = 10.836

Question What is the H+ concentration in the solution of pH 3.47 [H+] = 10-pH = 10 -3.4 = 3.39x10-4 mol/L or M

Questions What is the concentration OH- ions in a solution of pH of 10.47 Find concentration of H+ ions [H+] = 10-pH = 10-10.4 = 2.51x10-11 Find OH- concentration [H+ ] x [OH-] = 10-14 [OH-] = 10-14 2.51x10-11 = 3.98 x 10-4 mol/L or M

Find the pH of a 0.2 mol L-1 (0.2M) solution of H2SO4 Write the balanced equation for the dissociation of the acid H2SO4  2H+(aq) + SO42-(aq) Use the equation to find the [H+]: 0.2 mol L-1 H2SO4 produces 2 x 0.2 = 0.4 mol L-1 Calculate pH: pH = -log[H+] pH = -log [0.4] = 0.4

Pg 254 Q 9 a,b,c, 10 a,b 11