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Transcript
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: 1.donates a proton (H+) to a base. 2.Acids are proton donors. 3.When acids react with water, hydronium (H3O+) ions are produced. H+ ions cannot exist by themselves • HCl + H2O H3O+ + Clacid 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: 1. accepts a proton from an acid (Bases are proton acceptors. 2. 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 H N + H H + H Cl H N H - H + 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+ + NO3acid 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 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+). 1.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 H 2S HSNH4+ • Write the formulae of the conjugate acids of the following bases: • OHHCO3H 2O 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