ACID AND BASE By Dr. Taing N. You 2009 ~2010

Water Water behave both like acid and base The dissociation of water is the most fundamental of acid- base reaction Water break apart to form H+ and hydroxyl OH- In pure water, the equilibrium constant is defined by The equilibrium constant less then 1 suggest that the reaction prefers to stay on the side of the reactants.

Definition of ACID and BASE Arrhenius Acid: generate [H + ] in solution Base: generate [OH - ] in solution Normal Arrhenius equation: acid + base salt + water Example: HCl + NaOH NaCl + H 2 O Lewis Acid: Accept an electron pair Base: Donate an electron pair The advantage of this theory is that many more reaction can be considered acid-base reaction as they do not have to occur in solution

Definition of ACID and BASE Bronted –Lowery Acid: anything that donates a [H + ] Base: anything that accepts a [H + ] Normal Arrhenius equation: acid + base acid + base Example: HNO 2 + H 2 O NO H 3 O + Each acid has a conjugate base and each base has a conjugate acid.

pH and pOH The acidity or basicity of a substance is defined most typically by the pH, defined as follow: pOH gives another way to measure the acidity of a solution. It is just the opposite of pH pH + pOH = 14 pH = 7neutral (pH of pure water) pH < 7acid pH > 7 base

Salt A salt is formed when acid reacts with base Acid releases H + while base releases OH - Hydrolysis pH of the salt depends on the strengths of the original acids and bases Conjugate of a strong acid is very weak and cannot undergo hydrolysis and vice versa for conjugate of strong base acidBaseSalt pH Strong pH = 7 WeakStrongpH > 7 StrongWeakpH < 7 Weak Depends on which is stronger

Acid – Base character H-X is to be acid: Hydrogen must have positive number Hydrogen ionizes to form a positive +1 H-X is base when H have a negative charge NaH is not an acid because H has a -1 charge, its a base CH 4 is not an acid because H does not ionize H-O-X to be base: X-O-H must break to form OH - H-O-X is acid if H ionizes to form H +

Strong acid They are completely ionized in solution There are 7 strong acid: Hydrochloric acidHCl Bromic acid HBr Iodic acid HI Sulfuric acid H 2 SO 4 Nitric acidHNO 3 Chloric acidHClO 3 Perchloric acid HClO 4

Calculate pH of a strong acid It is easiest to follow the standard Start, Change, Equilibrium process Ex: Determine the pH of a 0.25 M sol of HBr HBr (aq) H + (aq)+Br - Start0.25 M0 M Change Equilibrium00.25

Weak acid They are the most common type of acid The equilibrium for dissociation of acid is know as K a The larger the value of K a the stronger the acid Example: Determine the pH of 0.3 M acetic acid with the K a of 1.8 x Write the equilibrium equation for acid:

Weak acid Write the equilibrium expression Start, Change, Equilibrium HC 2 H 3 O 2 (aq) H + (aq)+HC 2 H 3 O 2 - Start0.3 M0 M Change- x+ x Equilibrium0xx

Week acid Substitute the variable and solve for [H + ]

Strong base Like strong acid they completely ionize in solution There are 8 strong bases – Lithium hydroxideLiOH – Sodium hydroxideNaOH – Potassium hydroxideKOH – Rubidium(I) cation hydroxide RbOH – Cesium(I) cation hydroxide CsOH – Calcium hydroxide Ca(OH) 2 – Strontium dihydroxideSr(OH) 2 – Barium dihydroxide Ba(OH) 2

Calculate pH of a strong base It is easiest to follow the standard Start, Change, Equilibrium process Ex: Determine the pH of a 0.o10 M sol of Ba(OH) 2 Ba(OH) 2 (aq) Ba 2+ (aq) +2OH - (aq) Start0.10 M0 M Change Equilibrium00.10

Weak base This group follow the following equation K b is base dissociation constant:

Weak base The similar Start, Change, Equilibrium process is used to calculate the pH of weak base, however a few step is added Ex: Determine the pH of 0.15 M NH 3 with K b = 1.8 x Write the equilibrium expression and the K b value

Weak base Start, Change, Equilibrium process Substitute de variable and solve for [OH - ] NH 3 (aq) +H2OH2O NH 4 + +OH - Start0.15 M- 0 M Change- x- + x Equilibrium x- xx

Common ion effect When adding a salt to a weak acid or base that contains one of the ions present in the acid or base. Molarity of the salt must be added in the calculation of pH The same process is use to calculate K a and K b as in weak acid or weak base. Example: Find the pH of a solution formed by dissolving mol of HC 2 H 3 O 2 with a K a of 1.8x10 -8 and mol of NaC 2 H 3 O 2 in a total volume of 1.00 L.

Common ion effect HC 2 H 3 O 2 (aq) H + (aq)+HC 2 H 3 O 2 - Start0.10 M0 M0.20 M Change- x+ x Equilibrium xx0.2 + x

Acid Base titration: Strong acid/strong Base An acid-base titration is when you add a base to an acid until the equivalence point is reached which is where the moles of acid equal the moles of base. For the titration of a strong base and a strong acid, this equivalence point is reached when the pH of the solution is seven (7) as seen on the following titration curvetitration

Acid Base titration: Strong acid/strong Base The equivalence point is reached when the pH is greater than seven (7). The half equivalence point is when half of the total amount of base needed to neutralize the acid has been added. It is at this point where the pH = pK a of the weak acid.

Acid Base titration In an acid-base titration, the base will react with the weak acid and form a solution that contains the weak acid and its conjugate base until the acid is completely gone. To solve these types of problems, we will use the weak acid's K a value and the molarities in a similar way as we have before. Before demonstrating this way, let us first examine a short cut, called the Henderson-Hasselbalch Equation. This can only be used when you have some acid and some conjugate base in your solution.

Acid Base titration Where: pH: the log of the molar concentration of the hydrogen pKa: the equilibrium dissociation constant for an acid [base]: the molar concentration of a basic solution [acid]: the molar concentration of an acidic solution If you only have acid, then you must do a pure K a problem and if you only have base (like when the titration is complete) then you must do a K b problem.

Example Problem 25.0 mL of.400 M KOH is added to 100. mL of.150 M benzoic acid, HC 7 H 5 O 2 (Ka=6.3x10 -5 ). Determine the pH of the solution