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Bronsted-Lowry acids and bases

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.

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Bronsted-Lowry acids and bases

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  1. Bronsted-Lowry acids and bases

  2. 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.

  3. 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

  4. 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+

  5. 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.

  6. + H H - H N + H Cl N H H + Cl H H 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

  7. 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+

  8. 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

  9. Questions • Pg 245 Q 1,2,3

  10. 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+

  11. 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)

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

  13. 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

  14. 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.

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

  16. 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

  17. 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-

  18. 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-

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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 MpH = -log [H+]pH = -log 0.045pH = 1.35 • The pOH is given by pH + pOH = 14We substitute in the pH of 1.35 and get:1.35 + pOH = 14So, pOH = 12.65

  25. 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 Mhence, pOH = -log [OH-]pOH = -log 0.000685pOH = 3.164b) pOH is 3.164.The pH can be found by using pH + pOH = 14.Substituting in gives us pH + 3.164 = 14So, pH = 10.836

  26. 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

  27. 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

  28. 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

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

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