Advanced placement chemistry acids bases and aqueous equilibria
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ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA. Acids - taste sour Bases (alkali)- tastes bitter and feels slippery. Arrhenius concept - acids produce hydrogen ions in aqueous solution while bases produce hydroxide ions.

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Advanced placement chemistry acids bases and aqueous equilibria
ADVANCED PLACEMENT CHEMISTRYACIDS, BASES, AND AQUEOUS EQUILIBRIA


Acids taste sour bases alkali tastes bitter and feels slippery
Acids- taste sourBases(alkali)- tastes bitter and feels slippery


Arrhenius concept- acids produce hydrogen ions in aqueous solution while bases produce hydroxide ions


Bronsted lowry model acids are proton h donors and bases are proton acceptors
Bronsted-Lowry model- acids are proton (H+) donors and bases are proton acceptors


Lewis model acids are electron pair acceptors while bases are electron pair donors
Lewis model- acids are electron pair acceptors while bases are electron pair donors


hydronium ion (H3O+)- formed on reaction of a proton with a water molecule. H+ and H3O+ are used interchangeably in most situations.


HA(aq) + H2O(l) H3O+(aq) + A-(aq)Acid Base Conjugate Conjugate Acid Baseconjugate base- everything that remains of the acid molecule after a proton is lostconjugate acid- base plus a proton


Acid dissociation constant k a k a h 3 o a or k a h a ha ha
Acid dissociation constant (Ka)Ka = [H3O+][A-] or Ka = [H+][A-] [HA] [HA]


Strong acid - mostly dissociated - equilibrium lies far to the right - a strong acid yields a weak conjugate base (much weaker than H2O)Weak acid- mostly undissociated - equilibrium lies far to the left - has a strong conjugate base (stronger than water)


Common strong acids -all aqueous solutions (Know these!)H2SO4 (sulfuric)HCl (hydrochloric)HNO3 (nitric)HClO3 (chloric)HClO4 (perchloric)HI (hydroiodic)H2CrO4 (chromic)HMnO4 (permanganic)HBr (hydrobromic)


Sulfuric acid is a diprotic acid which means that it has two acidic protons. The first (H2SO4) is strong and the second (HSO4-) is weak.


Oxyacids- acidic protons. The first (H most acids are oxyacids - acidic proton is attached to OWeak oxyacids: H3PO4 (phosphoric) HNO2 (nitrous)HOCl (hypochlorous)


Within a series, acid strength increases with increasing numbers of oxygen atoms. For example: HClO4 > HClO3 > HClO2 > HClO and H2SO4 > H2SO3(Electronegative O draws electrons away from O-H bond)


Acid strength increases with increasing electronegativity of oxyacids. For example: HOCl>HOBr>HOI>HOCH3


Organic acids oxyacids. For example: HOCl>HOBr>HOI>HOCH- O have carboxyl group -C-OH - usually weak acidsCH3COOH acetic C6H5COOH benzoic


Hydrohalic acids oxyacids. For example: HOCl>HOBr>HOI>HOCH- H is attached to a halogen (HCl, HI, etc.)HF is the only weak hydrohalic acid. Although the H-F bond is very polar, the bond is so strong (due to the small F atom) that the acid does not completely dissociate.


Weak acid strength is compared by the K oxyacids. For example: HOCl>HOBr>HOI>HOCHa values of the acids. The smaller the Ka, the weaker the acid. Strong acids do not have Ka values because the [HA] is so small and can not be measured accurately.


Amphoteric substance oxyacids. For example: HOCl>HOBr>HOI>HOCH- Substance that can act as an acid or as a base. Ex. H2O, NH3, HSO4-(anything that can both accept and donate a proton)


autoionization of water oxyacids. For example: HOCl>HOBr>HOI>HOCH H2O + H2O  H3O+ + OH- base acid conjugate conjugate acid base


Ion product constant for water (K oxyacids. For example: HOCl>HOBr>HOI>HOCHw) Kw = [H3O+][OH-] Kw = [H+][OH-]At 25oC, Kw = 1 x 10-14 mol2/L2 because [H+] = [OH-] = 1 x 10-7 M


No matter what an aqueous solution contains, at 25 oxyacids. For example: HOCl>HOBr>HOI>HOCHoC, [H+][OH-] = 1 x 10-14 Neutral solution [H+] = [OH-] Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-]Kw varies with temperature


Ph log h if h 1 0 x 10 7 m ph 7 00
pH = -log [H oxyacids. For example: HOCl>HOBr>HOI>HOCH+]If [H+] = 1.0 x 10-7 M, pH = 7.00


Significant figures in pH and other log values oxyacids. For example: HOCl>HOBr>HOI>HOCH: The number of decimal places in the log value should equal the number of significant digits in the original number (concentration).


pOH = -log [OH oxyacids. For example: HOCl>HOBr>HOI>HOCH-]pK = -log KpH and pOH are logarithmic functions. The pH changes by 1 for every power of 10 change in [H+]. pH decreases as [H+] increases.


Ph poh 14 h antilog ph oh antilog poh
pH + pOH = 14 oxyacids. For example: HOCl>HOBr>HOI>HOCH[H+] = antilog(-pH) [OH-] = antilog(-pOH)


Calculating pH of Strong Acid Solutions oxyacids. For example: HOCl>HOBr>HOI>HOCH Calculating pH of strong acid solutions is generally very simple. The pH is simply calculated by taking the negative logarithm of concentration of a monoprotic strong acid. For example, the pH of 0.1 M HCl is 1.0. However, if the acid concentration is less than 1.0 x 10-7, the water becomes the important source of [H+] and the pH is 7.00. The pH of an acidic solution can not be greater than 7 at 25oC!!!!!


Another exception is calculating the pH of a H oxyacids. For example: HOCl>HOBr>HOI>HOCH2SO4 solution that is more dilute than 1.0 M. At this concentration, the [H+] of the HSO4- must also be calculated.


Ex calculate the h and ph in a 1 0 m solution of hcl
Ex. Calculate the [H oxyacids. For example: HOCl>HOBr>HOI>HOCH+] and pH in a 1.0 M solution of HCl.

HCl is a strong monoprotic acid, therefore its concentration is equal to the hydrogen ion concentration.

[H+] = 1.0 M

pH = - log (1.0) = 0.00


Ex calculate the ph of 1 0 x 10 10 m hcl
Ex. Calculate the pH of oxyacids. For example: HOCl>HOBr>HOI>HOCH1.0 x 10-10 M HCl.

Since the [H+] is less than 1.0 x 10-7, the [H+] from the acid is negligible and the pH = 7.00


Calculating pH of Weak Acid Solutions oxyacids. For example: HOCl>HOBr>HOI>HOCH Calculating pH of weak acids involves setting up an equilibrium. Always start by writing the equation, setting up the acid equilibrium expression (Ka), defining initial concentrations, changes, and final concentrations in terms of X, substituting values and variables into the Ka expression and solving for X.


Ex. Calculate the pH of a 1.00 x 10 oxyacids. For example: HOCl>HOBr>HOI>HOCH-4 M solution of acetic acid. The Ka of acetic acid is 1.8 x 10-5HC2H3O2 H+ + C2H3O2-Ka = [H+][C2H3O2-] = 1.8 x 10-5 [HC2H3O2]


R oxyacids. For example: HOCl>HOBr>HOI>HOCHeaction HC2H3O2 H+ + C2H3O2-

Initial 1.00 x 10-4 0 0

Change -x +x +x

Equilibrium 1.00 x 10-4 - x x x

Often, the -x in a Ka expression

1.8 x 10-5 = (x)(x) can be treated as negligible. 1.00x10-4 - x

1.8 x 10-5 (x)(x) x = 4.2 x 10-5 1.00 x 10-4


When you assume that x is negligible, you must check the validity of this assumption. To be valid, x must be less than 5% of the number that it was to be subtracted from. In this example 4.2 x 10-5 is greater than 5% of 1.00 x 10-4. This means that the assumption that x was negligible was invalid and x must be solved for using the quadratic equation or the method of successive approximation.


Use of the quadratic equation: validity of this assumption. To be valid, x must be less than 5% of the number that it was to be subtracted from. In this example 4.2 x 10x2 + 1.8 x 10-5x - 1.8 x 10-9 = 0x = 3.5 x 10-5 and -5.2 x 10-5Since a concentration can not be negative, x= 3.5 x 10-5 M x = [H+] = 3.5 x 10-5 pH = -log 3.5 x 10-5 = 4.46


Another method which some people prefer is the method of successive approximations. In this method, you start out assuming that x is negligible, solve for x, and repeatedly plug your value of x into the equation again until you get the same value of x two successive times.


Using successive approximation for the previous example would go as follows: x = 4.2 x 10-5 x = 3.2 x 10-5 x = 3.5 x 10-5 x = 3.4 x 10-5 x = 3.4 x 10-5[H+] = 3.4 x 10-5pH = 4.47


or/ with a graphing calculator: would go as follows: (1.8 x 10-5 x 1.00 x 10-4) = 4.2 x 10-5 (not negl)((1.8 x 10-5)( 1.00 x 10-4 -ans)) = 3.2 x 10-5 =3.5 x 10-5 =3.4 x 10-5 =3.4 x 10-5

Use answer key on calculator for this!

Press Enter key repeatedly until you get the same answer each time


Calculating pH of polyprotic acids would go as follows: All polyprotic acids dissociate stepwise. Each dissociation has its own Ka value. As each H is removed, the remaining acid gets weaker and therefore has a smaller Ka. As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton.


Except for H would go as follows: 2SO4, polyprotic acids have Ka2 and Ka3 values so much weaker than their Ka1 value that the 2nd and 3rd (if applicable) dissociation can be ignored. The [H+] obtained from this 2nd and 3rd dissociation is negligible compared to the [H+] from the 1st dissociation. Because H2SO4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0 M. The quadratic equation is needed to work this type of problem.


Ex calculate the ph of a 1 00 x 10 2 m h 2 so 4 solution the k a of hso 4 is 1 2 x 10 2
Ex. Calculate the pH of a 1.00 x 10 would go as follows: -2 M H2SO4 solution. The Ka of HSO4- is 1.2 x 10-2

H2SO4 H + + HSO4-

Before 1.00 x 10-2 0 0

Change -1.00 x 10-2 +1.00 x 10-2 +1.00 x 10-2

After 0 1.00 x 10-2 1.00 x 10-2

Reaction HSO4- H+ + SO4-

Initial 1 x 10-2 1x 10-2 0

Change -x +x +x

Equil. 1 x 10-2 -x 1 x 10-2 +x x


K would go as follows: a = [H+][SO4-]= 1.2 x 10-2 [HSO4-]

1.2 x 10-2 = (1 x 10-2 + x)(x)

(1 x 10-2 -x)

Using the quadratic equation,

x = 4.52 x 10-3

[H+]= 1 x 10-2 + (4.52 x 10-3)

= 1.45 x 10-2

pH = 1.84


Determination of the pH of a Mixture of Weak Acids would go as follows: Only the acid with the largest Ka value will contribute an appreciable [H+]. Determine the pH based on this acid and ignore any others.


Determination of the Percent Dissociation of a Weak Acid would go as follows: % dissociation = amt. dissociated (mol/L) x100 initial concentration (mol/L) = final [H+] x 100initial [HA]


For a weak acid, percent dissociation (or ionization) increases as the acid becomes more dilute. Equilibrium shifts to the right.


BASES increases as the acid becomes more dilute. Equilibrium shifts to the right.The hydroxides of Group I and IIA metals are all strong bases. The Group IIA hydroxides are not very soluble. This property allows some of them to be used effectively as stomach antacids.


Ex calculate the oh h and ph of a 0 0100 m solution of naoh
Ex. Calculate the [OH increases as the acid becomes more dilute. Equilibrium shifts to the right.-], [H+], and pH of a 0.0100 M solution of NaOH.

NaOH is a strong base.

[OH-] = 0.0100 M

[H+] = 1 x 10-14/1 x 10-2 =1.00 x 10-12 M

pH = - log 1.00 x 10-12 = 12.000


Weak bases (bases without OH increases as the acid becomes more dilute. Equilibrium shifts to the right.-) react with water to produce a hydroxide ion. Common examples of weak bases are ammonia (NH3), methylamine (CH3NH2), and ethylamine (C2H5NH2).


B(aq) + H increases as the acid becomes more dilute. Equilibrium shifts to the right.2O(l) BH+(aq) + OH-(aq) base acid conjugate conjugate acid base NH3 + H2O  NH4+ + OH- base acid conjugate conjugate acid base The lone pair on N forms a bond with a H+. Most weak bases involve N.


Base dissociation constant increases as the acid becomes more dilute. Equilibrium shifts to the right. (Kb)Kb = [BH+][OH-] Kb = [NH4+][OH-] [B] [NH3]


Determination of the pH of a weak base is very similar to the determination of the pH of a weak acid. Follow the same steps. Remember, however, that x is the [OH-] and taking the negative log of x will give you the pOH and not the pH!


Ex calculate the oh and the ph for a 15 0 m nh 3 solution the k b for nh 3 is 1 8 x 10 5
Ex. Calculate the [OH the determination of the pH of a weak acid. Follow the same steps. Remember, however, that -] and the pH for a 15.0 M NH3 solution. The Kb for NH3 is 1.8 x 10-5.

Reaction NH3 + H2O  NH4+ + OH-

Initial 15.0 --- 0 0

Change -x --- +x +x

Equil 15.0-x --- x x

Kb = 1.8 x 10-5 = [NH4+][OH-] = x2 x2

[NH3] 15.0-x 15.0

x = 1.6 x 10-2 = [OH-]

pOH = -log 1.6 x 10-2 = 1.78

pH = 14-1.78 = 12.22


Determination of the ph of salts
Determination of the pH of Salts the determination of the pH of a weak acid. Follow the same steps. Remember, however, that


Neutral Salts the determination of the pH of a weak acid. Follow the same steps. Remember, however, that - Salts that are formed from the cation of a strong base and the anion from a strong acid form neutral solutions when dissolved in water. Ex. NaCl, KNO3


Acid Salts the determination of the pH of a weak acid. Follow the same steps. Remember, however, that - Salts that are formed from the cation of a weak base and the anion from a strong acid form acidic solutions when dissolved in water. Ex. NH4Cl The cation hydrolyzes the water molecule to produce hydronium ions and thus an acidic solution.NH4+ + H2O  H3O+ + NH3 strong acid weak base


Basic Salts the determination of the pH of a weak acid. Follow the same steps. Remember, however, that - Salts that are formed from the cation of a strong base and the anion from a weak acid form basic solutions when dissolved in water. Ex. NaC2H3O2, KNO2 The anion hydrolyzes the water molecule to produce hydroxide ions and thus a basic solution.C2H3O2- + H2O  OH- + HC2H3O2 strong base weak acid


When determining the exact pH of salt solutions, we can use the Ka of the weak acid formed to find the Kb of the salt or we can use the Kb of the weak base formed to find the Ka of the salt. Ka x Kb = Kw


Ex. Calculate the pH of a 0.15 M solution of sodium acetate. Sodium acetate is the salt of a strong base (NaOH) and a weak acid (acetic acid) and thus forms a basic solution. The acetate ion hydrolyzes to produce acetic acid and hydroxide ions.

Reaction C2H3O2- + H2O  HC2H3O2 + OH-

Initial 0.15M - 0 0

Change -x +x +x

Equil. 0.15- x x x


K acetate.b =[HC2H3O2][OH-] [C2H3O2-]

Kb = Kw = 1 x 10-14 = 5.6 x 10-10 Ka 1.8 x 10-5

5.6 x 10-10 = x2 x2 x = 9.2 x 10-6 0.15 - x 0.15

[OH-] = 9.2 x 10-6

pOH = - log 9.2 x 10-6 = 5.04

pH = 14.00-5.04 = 8.96


Acidic and Basic Oxides acetate.When metallic (ionic) oxides dissolve in water they produce a metallic hydroxide (basic solution). When nonmetallic (covalent) oxides dissolve in water they produce a weak acid (acidic solution). CaO + H2O  Ca(OH)2 CO2 + H2O  H2CO3


Salts of Highly Charged Metals acetate. Salts that contain a highly charged metal ion produce an acidic solution.AlCl3 + 6H2O  Al(H2O)63+ + 3Cl- Al(H2O)63+Al(H2O)5(OH) 2+ + H+


The higher the charge on the metal ion the stronger the acidity of the hydrated ion. The electrons are pulled away from the O-H bond and toward the positively charged metal ion. FeCl3 and Al(NO3)3 also behave this way.


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