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Acid-Base Equilibria and Solubility Equilibria. Chapter 16 16.1-16.9. Acid-Base Equilibria and Solubility Equilibria. We will continue to study acid-base reactions Buffers Titrations We will also look at properties of slightly soluble compunds and their ions in solution.
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Acid-Base Equilibria and Solubility Equilibria Chapter 16 16.1-16.9
Acid-Base Equilibria and Solubility Equilibria • We will continue to study acid-base reactions • Buffers • Titrations • We will also look at properties of slightly soluble compunds and their ions in solution
Homogeneous vs. Heterogeneous • Homogeneous Solution Equilibria- a solution that has the same composition throughout after equilibrium has been reached. • Heterogeneous Solution Equilibria- a solution that after equilibrium has been reached, results in components in more than one phase.
The Common Ion Effect • Acid-Base Solutions • Common Ion • The Common Ion Effect- the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. • pH
The Common Ion Effect The presence of a common ion suppresses the ionization of a weak acid or a weak base. CH3COONa (s) Na+(aq) + CH3COO-(aq) common ion CH3COOH (aq) H+(aq) + CH3COO-(aq) Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid).
[conjugate base] pH = pKa + log [acid] Henderson-Hasselbach Equation • Relationship between pKa and Ka. pKa = -log Ka Henderson-Hasselbalch equation
HCOOH (aq) H+(aq) + HCOO-(aq) Initial (M) Change (M) Equilibrium (M) pH Calculations What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH pKa = 3.77 Mixture of weak acid and conjugate base! 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x
pH Calculations [HCOO-] pH = pKa + log [HCOOH] [0.52] pH = 3.77 + log [0.30] Common ion effect 0.30 – x 0.30 0.52 + x 0.52 = 4.01
Buffer Solutions • A buffer solution is a solution of: • A weak acid or a weak base and • The salt of the weak acid or weak base • Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.
Buffer Solutions • IV solutions (~7.4) • Blood (~7.4) • Gastric Juice (~1.5)
Buffer Solutions Consider an equal molar mixture of CH3COOH and CH3COONa H+(aq) + CH3COO-(aq) CH3COOH (aq) OH-(aq) + CH3COOH (aq) CH3COO-(aq) + H2O (l) Add strong acid Add strong base
Preparing a Buffer Solution with a Specific pH • Work Backwards • Choose a weak acid whose pKa is close to the desired pH • Substitute pKa and pH values into the Henderson-Hasselbach Equation • This will give a ratio of [conjugate Base]/[Acid] • Convert ratio to molar quantities
Acid-Base Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Types of Titrations Those involving a strong acid and a strong base Those involving a weak acid and a strong base Those involving a strong acid and a weak base
Acid-Base Titrations • Indicator – substance that changes color at (or near) the equivalence point Equivalence point – the point at which the reaction is complete Slowly add base to unknown acid UNTIL The indicator changes color (pink)
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) OH-(aq) + H+(aq) H2O (l) Strong Acid-Strong Base Titrations Strong Acid-Strong Base Titrations
Strong Acid-Base Titrations • Calculate the pH after the addition of 10.0mL of 0.100M NaOH to 25.0mL of 0.100M HCl. # moles NaOH 10.0mL x (0.100mol NaOH/ 1L NaOH) x (1L/ 1000mL) = 1.00 x 10-3 mol #moles HCl 25.0mL x (0.100mol HCl/ 1 L HCl) x (1L/ 1000mL) = 2.50 x 10-3 mol Amount of HCl left after partial neutralization 2.50 x 10-3 mol - 1.00 x 10-3 mol = 1.5 x x 10-3 mol
Strong Acid-Base Titrations Thus, [H+] = 1.5 x x 10-3 mol/ 0.035L [H+] = 0.0429 M pH = -log [H+] pH = -log 0.0429 pH = 1.37
Strong Acid-Base Titrations • Calculate pH after the addition of 35.0mL of 0.100M NaOH to 25.0mL of 0.100M HCl. # moles NaOH 0.100 mol NaOH / 0.035 L NaOH = 3.50 x 10-3 mol #moles HCl 0.100 mol HCl / 0.025 L HCl = 2.50 x 10-3 mol Amount of NaOH left after full HCl neutralization 3.50 x 10-3 mol – 2.50 x 10-3 mol = 1.0 x x 10-3 mol
Strong Acid-Base Titrations Thus, [NaOH] = 1.0 x x 10-3 mol/ 0.06L [NaOH] = 0.0167 M [OH-] = 0.0167 M pOH = -log [H+] pOH = -log 0.0167 pOH = 1.78 pH = 14.0-pOH pH = 14.0-1.78 pH = 12.22
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH-(aq) CH3COO-(aq) + H2O (l) CH3COO-(aq) + H2O (l) OH-(aq) + CH3COOH (aq) Weak Acid-Strong Base Titrations Weak Acid-Strong Base Titrations At equivalence point (pH > 7):
HCl (aq) + NH3(aq) NH4Cl (aq) H+(aq) + NH3(aq) NH4Cl (aq) NH4+(aq) + H2O (l) NH3(aq) + H+(aq) Strong Acid-Weak Base Titrations Strong Acid-Weak Base Titrations At equivalence point (pH < 7):
Acid-Base Indicators • Equivalence point occurs when OH- = H+ originally present. • Indicators • End Point- Occurs when indicator changes color • End point ~ Equivalence point
pH Acid-Base Indicators
Solubility Equilibria • Reactions that produce precipitates • Importance • Tooth Enamel + Acid = tooth decay • Barium Sulfate = used in x-rays • Fudge!!!
AgCl (s) Ag+(aq) + Cl-(aq) Solubility Equilibria • Solubility Product Constant (Ksp)- the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. Ksp= [Ag+][Cl-]
MgF2(s) Mg2+(aq) + 2F-(aq) Ag2CO3(s) 2Ag+(aq) + CO32-(aq) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq) Solubility Products • What are the correct solubility products of the following equations? Ksp= [Mg2+][F-]2 Ksp= [Ag+]2[CO32-] Ksp= [Ca2+]3[PO33-]2
Solubility Constants • What does a large Ksp mean? • What does a small value mean?
Molar Solubility and Solubility Molar solubility(mol/L)- is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility(g/L)- is the number of grams of solute dissolved in 1 L of a saturated solution.