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Ionic Equilibrium in Solutions

Ionic Equilibrium in Solutions. K sp , K a and K b. Ionic Equilibrium. Much like with a system of equations, a solution is also an equilibrium NaCl ( aq )  Na + ( aq ) + Cl - ( aq ) The ions in this solution are constantly dissociating and re-associating. Ionic Equilibrium.

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Ionic Equilibrium in Solutions

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  1. Ionic Equilibrium in Solutions Ksp, Ka and Kb

  2. Ionic Equilibrium • Much like with a system of equations, a solution is also an equilibrium • NaCl(aq)  Na+(aq) + Cl-(aq) • The ions in this solution are constantly dissociating and re-associating

  3. Ionic Equilibrium • What is a saturated solution? • A solution which has reached its capacity of a solute • What is a super-saturated solution? • A solution which holds more than its full capacity of a solute • Video demonstration… • Now onto the real stuff…

  4. What we will be covering... • Ka – The acidity constant • Kb – The Alkalinity (base) constant • Ksp – Solubility product constant • Kwater – Water ionization constant

  5. Arrhenius theory of acids and bases • An acid is a substance that dissociates in water to produce hydrogen ions (H+) • HCl(aq) -> H+(aq) + Cl-(aq) • A base is a substance that dissociates in water to produce hydroxide ions (OH-) • NaOH(aq) -> Na+(aq) + OH-(aq)

  6. Neutralization • Acids are neutralized by a base and vice versa • NaOH + HCl -> NaCl + H2O • Acids and bases can be stronger or weaker • You need more of a weak base to neutralize a strong acid

  7. Acid or base? • NaOH • Base! • HCl • Acid! • H2SO4 • Acid! • NH3 • Base!

  8. Brønsted-Lowry theory of acids and bases • An acid is a substance in which a proton (Hydrogen atom, H+) can be removed. An acid is seen as a proton donor. Seeing how a single H+ cannot exist on its own, it can also be shown as a hydronium ion (H3O+) • A base is a substance that can remove a proton from an acid. A base is seen as a proton acceptor.

  9. Conjugate Acid-Base Pairs • In each acid/base reaction, there are 2 conjugate acid-base pairs • Ex: HCl (aq) + H2O (aq) -> H3O+ (aq)+ Cl- (aq) • HCl and Cl- are one pair (A-B) • HCl is the acid and Cl- is the base • H2O and H3O+ are the other pair • H2O is the base and H3O+ is the acid

  10. Conjugate Acid-Base Pairs • NH3 (aq) + H2O (aq) -> NH4+(aq) + OH-(aq) • NH3 and NH4+ • NH3 is the base and NH4+ is the acid • H2O and OH- • H2O is the acid and OH- is the base

  11. Water, Acid or Base? • Looking back at the two conjugate Acid-Base pairs, is H2O an acid or a base? • In the Brønsted-Lowry theory of acids and bases, water can be considered an acid or a base, depending on its role in the reaction.

  12. Ionization constant of Water • The dissociation of ions in water is an equilibrium • The pH of the solution, which measures its acidity, is determined by where the equilibrium settles • This equilibrium can be quantified using the ionization of water constant Kwater

  13. Ionization constant of Water • This constant Kwater makes it possible to understand the interdependence of Hydronium ions (H3O+) and Hydroxide ions (OH-) • Before we explore this wonderful relationship, let us go over what exactly the pH and pOH are...

  14. What are pH and pOH? • pH is a quantitative value attributed to the acidity of a solution • The lower the pH value, the higher the concentration of the hydronium ions (H3O+), and therefore the stronger the acid. • pOH is a quantitative value attributed to the alkalinity or basicity of a solution • The lower the pOH, the higher the concentration of the hydroxide ions (OH-) and therefore the stronger the base.

  15. Mathematical Expressions • pH and pOH can be expressed as the following mathematical expressions • pH = -log [H3O+] • [H3O+] = 10-pH • pOH = -log [OH-] • [OH-] = 10-pOH

  16. Example #1 • Express in the form of pH, the hydronium (H3O+) concentration of 4.7 x 10-11 mol/L in an aqueous solution. Is this solution acidic, neutral or basic? • Data • [H3O+] = 4.7 x 10-11 • pH = ?

  17. Solution • pH = - log [H3O+] • pH = - log (4.7 x 10 -11) • pH = 10.33 • The solution is basic due to its pH being higher than 7

  18. Example 2 • Express the pOH of 3.60 in the form of the hydroxide (OH-) concentration • Data • pOH = 3.60 • [OH-] = ?

  19. Solution • [OH-] = 10 -pOH • [OH-] = 10 -3.60 • [OH-] = 2.5 x 10 -4 • The concentration of hydroxide (OH-) is 2.5 x 10 -4 mol/L

  20. Relationship between pH and pOH • What is [H3O+] at pH 7? • 1.00 x 10 -7 • What is [OH-] at pOH 7? • 1.00 x 10 -7

  21. Ionization Constant of Water • The ionization of water follows this simple formula • 2 H2O (l)  H3O+ (aq) + OH- (aq) • Once this equation has reached equilibrium, we obtain the ionization of water constant Kwater

  22. Kw • Kw = [H3O+] x [OH-] • If water is neutral, pH 7, then we know the concentrations of the hydronium and hydroxide ions. • The concentration of both ions is 1.00 x 10-7 • Therefore...

  23. Ionization constant of water • Kw = [H3O+] x [OH-] • Kw = 1.00 x 10-7 x 1.00 x 10-7 • Kw = 1.00 x 10-14 • This is always at 25°C

  24. How this all fits together... • By carrying out the logarithmic inverse of each side of the equation, the following equivalence can be obtained • -log [H3O+] + -log [OH-] = -log (1.00 x 10-14) • -log [H3O+] + -log [OH-] =14 • pH + pOH = 14

  25. Relationship between the pH and [H3O+] and [OH-] • Knowing that Kw is constant in all aqueous solutions, we can use this to determine the concentration H3O+ and OH- ions in any acidic or basic solutions • Example! • At 25°C, a hydrochloric acid solution has a pH of 3.2. What is the concentration of each of the ions in this solution?

  26. Solution • [H3O+] = 10 –pH • [H3O+] = 10 -3.2 = 6.3 x 10-4 • Kw = [H3O+] x [OH-] = 1.00 x 10-14 • [OH-] = 1.00 x 10-14 / [H3O+] • [OH-] = 1.00 x 10-14 / 6.3 x 10-4 • [OH-] = 1.58 x 10-11

  27. Ka and Kb Acidity and Basicity Constants

  28. Ka and Kb • Here we will be quantifying the strength of acids and bases • The stronger the acid or base depends on how it dissociates • The more dissociation, the stronger the acid

  29. Strength of Acids • When an acid comes into contact with water, a certain amount of dissociation takes place • In a strong acid, as much as 100% will dissociate • Ex: HCl • In a weak acid, very little will dissociate. As little as 1% • Ex: Acetic acid (Vinegar)

  30. Acid Dissociation

  31. Ionization Percentage • Ionization percentage can be calculated by dividing the concentration of the H3O+ ions by the concentration of the original acid and multiplied by 100 • % = [H3O+] eq / [HA] i * 100 • Ensure that all of the concentrations are in the same units

  32. Calculating the Acidity • This can only be done using a weak acid, why? • If there is none of the original acid left, you can’t calculate an equilibrium constant • Using the following general equilibrium, we can calculate the Ka • HA (aq) + H2O (l) H3O+(aq) + A-(aq)

  33. Acidity Constant • Ka = ([H3O+] * [A-]) / [HA] • Since water is a liquid, there is no concentration • We cannot do this with a strong acid because there is none of the original HA acid left and you cannot divide by 0 • Would a weak acid have a higher or lower acidity constant? • Lower!

  34. Acidity Constant • To find the acidity constant, all of the concentrations must be known • Also, if the Ka is known, then we can use that to predict either the final concentration of the [H3O+], or the initial concentration of the [HA] • It can also be used to calculate the pH

  35. Basicity Constant • The basicity constant can also be calculated along the same lines • Using the following general equilibrium formula • B (aq) + H2O (l) HB+(aq) + OH-(aq) • Kb = ([HB+] * [OH-]) / [B] • Again, the weaker the base, the smaller the constant

  36. Ksp Solubility Product Constant

  37. Solubility Product Constant • A saturated solution that contains non-dissolved solute deposited at the bottom of a container is an example of a system at equilibrium. • The solubility of a substance corresponds to the maximum quantity of a substance that dissolves in a given volume of water • Usually given as g/100ml

  38. Example • BaSO4(s) Ba2+(aq) + SO42-(aq) • Ksp = [Ba2+] * [SO42-] • General formula • XnYm(s)  nX+(aq) + mY-(aq) • Ksp = [X+]n * [Y-]m

  39. Problem • The solubility of silver carbonate (Ag2CO3) is 3.6 x 10-3 g/100ml of solvent at 25°C. Calculate the value of the solubility product constant of silver carbonate. • Steps • 1- Find the concentration of the Ag2CO3 using M = m / n then the solubility • Where M is molar mass, m is mass and n is the amount in moles of Ag2CO3 • 2- Calculate the Ksp

  40. Solution • 1- M = m/n • n = m/M • n = 3.6 x 10-3 g / 275.8 g/mol • n = 1.3 x 10-5 mol for 100 ml (0.1 L) • Solubility = 1.3 x 10-5 / 0.1 L • Solubility = 1.3 x 10-4 mol/L

  41. Solution Continued • Ag2CO3 2 Ag+(aq) + CO32-(aq) • Ksp = [X+]n * [Y-]m • Ksp = [Ag+]2 * [CO32-] • [Ag+] = 2 * [Ag2CO3] = 2 * 1.3 x 10-4 mol/L • [Ag+] = 2.6 x 10-4 mol/L • [CO32-] = [Ag2CO3] = 1.3 x 10-4 mol/L • Ksp = [Ag+] * [CO32-] • Ksp = (2.6 x 10-4)2 mol/L * 1.3 x10-4 mol/L • Ksp = 8.8 x 10-12

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