"You can dance anywhere, even if only in your heart." ~Unknown "If dancing were any easier it would be called football." ~anonymous

MORE ACIDS AND BASES CONJUGATE ACID-BASE PAIR – TWO SUBSTANCES RELATED TO EACH OTHER BY DONATING AND ACCEPTING A SINGLE PROTON.* HA + H 2 O == H 3 O + + A - THIS EQUATION REPRESENTS A COMPETITION BETWEEN TWO BASES FOR A PROTON. K a = [H 3 O + ][A - ]/[HA] = [H + ][A - ]/[HA] K a = ACID DISSOCIATION CONSTANT BOTH H 3 O + AND H + ARE USED TO REPRESENT THE HYDRATED PROTON. [H 2 O] IS NOT INCLUDED IN THE EXPRESSION SINCE WATER DOES NOT AFFECT THE EQUILIBRIUM EXPRESSION. *ZUMDAHL

WATER PLAYS A VERY IMPORTANT ROLE IN THE IONIZATION OF AN ACID. STRENGTH OF AN ACID IS DEFINED BY ITS EQUILIBRIUM POSITION IN THE DISSOCIATION REACTION. FOR A STRONG ACID, THE EQUILIBRIUM POSITION IS FAR TO THE RIGHT. HA (aq) + H 2 O (aq) == H 3 O + (aq) + A - (aq) THERE IS A RELATIONSHIP BETWEEN THE STRENGTH OF AN ACID AND THAT OF ITS CONJUGATE BASE. A STRONG ACID HAS A WEAK CONJUGATE BASE.

PROPERTYSTRONG ACIDWEAK ACID K a valueK a is largeK a is small Position of equilibrium in dissociation Far to rightFar to left Equilibrium conc. of [H + ] compared to [HA] [H + ] >> [HA][H + ] << [HA] Strength of conjugate base compared to H 2 O A - much weaker than H 2 O A - much stronger than H 2 O

THE STRENGTH OF AN ACID IS INVERSELY RELATED TO THE STRENGTH OF ITS CONJUGATE BASE. CONVERSELY, THE STRENGTH OF A BASE IS INVERSELY RELATED TO THE STRENGTH OF ITS CONJUGATE ACID. AN AMPHOTERIC SUBSTANCE IS A SUBSTANCE THAT CAN BEHAVE AS AN ACID OR A BASE. WATER IS THE MOST COMMON EXAMPLE. H 2 O (acid 1) + H 2 O (base 1) == H 3 O + (acid 2) + OH - (base 2) K W = [H 3 O + ][OH - ] = [H + ][OH - ] = ion product constant for water [H +] = [OH -] = 1 x M at 25 o C (by experiment) K w = (1 x )(1 x ) = 1 x

IMPORTANT!! IN ANY AQUEOUS SOLUTION, NO MATTER WHAT IT CONTAINS, THE PRODUCT OF [H + ] AND [OH - ] MUST ALWAYS EQUAL 1 X AT 25 o C. THERE ARE THREE POSSIBLE SOLUTIONS: NEUTRAL SOLUTION[H + ] = [OH - ] ACIDIC SOLUTION[H + ] > [OH - ] BASIC SOLUTIION[H +] < [OH - ] BUT, IN EACH CASE, K w = [H + ][OH - ] = 1 X

COMMENT ON SIGNIFICANT FIGURES: THE NUMBER OF DECIMAL PLACES IN THE LOG IS EQUAL TO THE NUMBER OF SIGNIFICANT FIGURES IN THE ORIGINAL NUMBER. [H + ] = 1.0 X M TWO SIG. FIGURES pH = - log [H + ] = 9.00 Two decimal places pH IS A LOG SCALE. FOR EVERY UNIT CHANGE, [H + ] CHANGES BY A FACTOR OF 10.

pH + pOH = 14 EXAMPLE: THE pH OF HUMAN BLOOD IS 7.41 AT 25 o C. CALCULATE THE pOH, [H + ], AND [OH - ]. pH = 7.41 = -log [H + ] Log [H + ] = [H + ] = antilog (-7.41) = = 3.9 x M pOH = 14 – 7.41 = 6.59 [OH - ] = antilog (-pOH) = antilog (-6.59) = 2.6 x M

STRONG ACIDS AND BASES STRONG ACIDS AND BASES ARE, FOR ALL PRACTICAL PURPOSES, COMPLETELY DISSOCIATED (IONIZED). FOR EXAMPLE, IN A SOLUTION OF 1 M HCl, THERE WOULD BE NO MEASURABLE [HCl]. IT WOULD CONTAIN 1 M [H + ] AND 1 M [Cl - ]. H 2 O COULD ALSO FURNISH H + THROUGH AUTOIONIZATION, BUT THE AMOUNT WOULD BE INSIGNIFICANT COMPRED TO THE AMOUNT FROM HCl. THE AUTOIONIZATION REACTION WOULD BE PUSHED TO THE LEFT BECAUSE OF THE H + FROM HCl. H 2 O (l) == H + (aq) + Cl - (aq) pH = - log [H + ] = - log (1.0) = 0

EXAMPLE: CALCULATE THE pH OF A 0.10 M HNO 3 SOLUTION. SOURCES OF H + : 1)H + FROM HNO 3 (0.10 M) 2)H + FROM H 2 0 ( WOULD BE VERY SMALL) [H + ] = 0.10 M pH = -log (0.1) = 1.00 CALCULATE THE pH OF A 1.0 X M HCl SOLUTION. SOURCES OF H + : 1)FROM HCl (1 X ) ( WOULD BE SMALL) 2)FROM H 2 O (1 X ) SO pH ~ 7

WEAK ACIDS CALCULATE THE pH OF A 1.00 M SOLUTION OF HF (K a = 7.2 X ) MAJOR SPECIES IN SOLUTIO WOULD BE: 1)HF 2)H 2 O BOTH CAN FURNISH H + HF (aq) == H + (aq) + F - (aq) K a = 7.2 x H 2 O (l) == H + (aq) + F - (aq) K w = 1 x K a >> K w SO HF WILL BE PRIMARY SOURCE OF H + K a = [H + ][F - ]/[HF] = 7.2 x [H + ] = [F - ] = x [HF] = 1.00 – x K a = x 2 /(1.00 – x) if x << 1.00, then K a = x 2 /1.00 X 2 = 7.2 x and x = 2.7 x M pH = - log (2.7 x ) pH = 1.57

CALCULATE THE pH OF A M SOLUTION OF HYPOCHLOROUS ACID, HOCl, K a = 3.5 x MAJOR SPECIES: 1)HOCl K a = 3.5 X )H 2 OK w = 1 x HOCl IS STRONGER THAN WATER, AS K a >> K w. SO IT WILL DOMINATE H + PRODUCTION. AT EQUILIBRIUM, [HOCl] = – X AND [H + ] = [OCl - ] = X K a = X 2 /(0.100 – X) = 3.5 X K a IS SMALL, SO >> X AND K a = X 2 /0.100 X = 5.9 X M = [H + ] pH = - log (5.9 x ) = 4.23

WE TREAT STRONG BASES THE SAME WAY WE TREATED STRONG ACIDS. A BASIC SOLUTION AT 25 o C HAS A pH > 7. THE HYDROXIDES OF THE ALKALI METALS ARE STRONG BASES. THE HYDROXIDES OF ALUMINUM AND THE HEAVIER ALKALINE EARTH METALS (Ca, Sr, Mg, Ba) ARE STRONG BASES, BUT THEY HAVE LIMITED SOLUBILITY. THIS CAN BE USED TO ADVANTAGE. Al(OH) 3 AND Mg(OH) 2 ARE USED AS ANTACIDS. THE LOW SOLUBILITY PREVENTS HIGH OH - CONCENTRATIONS AND DAMAGE TO TISSUES. SLAKED LIME, Ca(OH) 2, IS USED TO SCRUB STACK GASES OF SO 2.

CALCULATE THE pH OF A 5.0 X M NaOH SOLUTION. MAJOR SPECIES: 1)Na + 2)OH - 3)H 2 O [OH - ] = 5.0 X K w = [H + ][OH - ] SO [H + ] = K w /[OH - ] = (1 X )/(5.0 X ) [H + ] = 2.0 X AND pH = - log (2 x ) = 12.70

A BASE DOES NOT HAVE TO CONTAIN HYDROXIDE ION. NH 3 (aq) + H 2 O (l) == NH 4 + (aq) + OH - (aq) BASES SUCH AS AMMONIA HAVE AN UNSHARED PAIR OF ELECTRONS ON THE NITROGEN THAT CAN ACCEPT A PROTON. B (aq) + H 2 O (l) == BH + (aq) + OH - (aq) K b = [BH + ][OH - ]/[B]

STRONG BASES HAVE VERY LARGE K b VALUES. WEAK BASES HAVE RELATIVELY SMALL K b VALUES. NH 3 (aq) + H 2 O (aq) == NH 4 + (aq) + OH - (aq) K b = [NH 4 + ][OH - ]/[NH 3 ] = 1.8 x EXAMPLE: CALCULATE THE pH OF A 15.0 M NH 3 SOLUTION. [NH 4 + ] = [OH - ] = X AND [NH 3 ] = 15 – X ~ 15 K b = x 2 /15 = 1.8 x x = 1.6 x M = [OH - ] [H + ] = K w /[OH - ] = (1 x )/(1.6 x ) [H + ] = 6.3 x pH = - log (6.3 x ) = 12.20