1. Goes with chapter 18 – Silberberg: Principles of General Chemistry Mrs. Laura Peck, 2013 1

2.  Understand the acid-base theories of Arrhenius, Bronsted- Lowry, and Lewis  Identify strong acids and bases and calculate their pH’s  Calculate the pH of a weak acid or base  Calculate the [] of a strong or weak acid or base from its pH  Predict the pH of a salt from its formula and then calculate the pH of the salt  Identify the components of a buffer and perform calculations involving the preparation of a buffer and the addition of strong acid or strong base to a buffer  Perform calculations involving strong acid-strong base titrations as well as weak acid-strong base and weak base- strong acid calculations.  Be familiar with titatration curves and selection of an acid- base indicator 2

3.  Arrhenius Theory  States that, in aqueous solution (water), acids produce hydrogen ions and bases produce hydroxide ions.  Bronsted-Lowry theory  Says that an acid is a proton (H + ) donor and a base is a proton acceptor  In the reaction below, HNO 3 transfers a proton to H 2 O forming H 3 O +, the hydronium ion.  H 3 O + is the conjugate acid of H 2 O and NO 3 - is the conjugate base of HNO 3.  The formulas in a conjugate acid-base pair differ by one H + HNO 3 + H 2 O  H 3 O + + NO 3 - Acid base c.acid c.base  Lewis Acid-base model  A lewis acid is an electron pair acceptor and a Lewis base is an electron pair donor.  Remember that this is a much broader definition than the other two, in that the acid-base interaction does not need to involve a transfer of a proton H+. 3

4.  Give the formulas for the conjugate base of H 2 SO 4 and the conjugate acid of CH 3 NH 2 4 HSO 4 - is the conjugate base of H 2 SO 4 CH 3 NH 3 + is the conjugate acid of CH 3 NH 2 **Note that each conjugate acid-base pair differs by 1H + HSO 4 - + 1H + = H 2 SO 4 ; CH 3 NH 3 + = 1H + + CH 3 NH 2

5.  The names and formulas of the six strong acids must be memorized.  The six strong acids are: HCl, HI, HNO 3, H 2 SO 4, and HClO 4  If an acid is not one of the six in the list, then for purposes of the AP exam, you can assume it is a weak acid.  The chart below compares the dissociation of strong acids to weak acids. 5 Type of acid, HAReversibility of reaction Ka valueIons existing when acid, HA, dissociates in water StrongNot reversibleKa value very large H+ and A- only. No HA present WeakReversibleKa is smallH+, A-, and HA

6.  Weak acids exist in equilibrium with their ions in aqueous solution.  The acid dissociation constant, K a, measures the extent to which the acid dissociates in water:  HA (aq) + H 2 O (l)   H 3 O + (aq) + A - (aq)  The equilibrium expression for the reaction is  K a = [H 3 O + ][A - ]  [HA]  A table of Ka values for monoprotic acids, containing one acidic hydrogen, appears in the appendix of your textbook and is worth studying.  The larger the K a value, the stronger the acid. 6

7.  Example #2: List the acids in order of increasing strength: HCN, HCl, HClO 2, HNO 2  Example #3: Arrange the following species in order of increasing base strength: NO 2 -, ClO 2 -, CN -, Cl - 7 HCN, HNO 2, HClO 2, and HCl. The first three weak acids are listed In order of increasing Ka values. HCl is stronger than all of the Weak acids given. HCl is a strong acid Cl -, ClO 2 -, NO 2 -, CN - ; the bases are listed in reverse order of Their conjugate acids in the previous example because the Stronger and acid, the weaker its conjugate base.

8.  Strong bases include groups I and II hydroxides such as NaOH.  Weak bases such as NH 3, are not group I or II hydroxides.  The base dissociation constant, K b, measures the extent to which a base reacts with water.  The reaction of a weak base, B, with water and its corresponding equilibrium expression is:  B (aq) + HOH (l)   BH + (aq) + OH - (aq) K b = [BH + ][OH - ] [B]  **AP tip: Writing the reaction for base dissociation can be tricky. Always remember to react the base with water. Use the Arrhenius and Bronsted- Lowry theories to help write the reaction. The base must react with water to produce hydroxide ions according to Arrhenius, and accepts an H+ according to Bronsted-Lowry. Check the charges of the reactant and products and be sure that the sums of the charges on both sides of the reaction are equal. 8

9.  For binary acids, HX, the strength of the H-X bond and the polarity of the bond will determine the behavior of the acid.  The polarity of the bonds in hydrogen halides become less polar going down a group.  The strength of the H-F bond is what makes it a weak acid, whereas the rest of the hydrogen halides are strong acids: HI>HBr>HCl  For a given series of Oxoacids such as HClO 4, HClO 3, HClO 2, and HClO, the acid strength increases with increasing number of oxygens attached to the central atom.  HClO 4 is a strong acid.  The remainder of the oxoacids are listed in order of decreasing strength (decreasing number of oxygen atoms)  The O-H bond becomes more polarized and weakened due to the electron density drawn toward the highly electronegative oxygen atoms. 9

10.  The pH of a strong acid can be calculated directly from the hydrogen ion concentration  pH = -log[H+]  [H+]. The molar concentration of the hydrogen ion, is obtained from the molarity of the acid.  The pH of a strong base can be calculated from its hydroxide ion concentration.  pOH = -log[OH-]  pOH + pH = 14.00  The concentration of a strong acid or strong base can be determined from the solution’s pH. 10

11.  Example #4: Calculate the pH of 0.010M HCl  Example #5: The pH of a Sr(OH) 2 solution is 13.50. Calculate the concentration of Sr(OH) 2  **AP tip: the number of sig figs in a pH measurement is equal to the number of decimal places in the pH. For example, 1.70 has 2 sig figs. 11 The pH equals 2.00; -log(0.010) = 2.00 pOH = 14 – pH  14 – 13.5 = 0.50 [OH] = inv log(-pOH)  inv log (-0.50) = 0.32M OH- 0.32 mol OH x 1 mol Sr(OH) 2 = 0.16M Sr(OH) 2 1 L 2 mol OH-

12.  The pH of a weak acid cannot be calculated directly from the concentration of the acid since all of the acid does not dissociate to form H+.  The equilibrium reaction of the acid must be considered.  Example #6: Calculate the pH of 0.25M HCN 12 First, write the reaction of the acid with water. Use the Arrhenius and Bronsted-Lowry theories to help you write the products. Check that you Have the correct charges on the products. HCN + H 2 O   H 3 O + + CN - now, set up an ‘ICE’ chart for equilibrium I.0.25 0 0 C.-x +x +x E. 0.25-x x x Third, write the equilibrium expression for Ka in the same manner as you did in the Last topic. Plug in the values from the Equilibrium line of the ICE chart. K a = 6.2x10 -10 = [H 3 O + ][CN - ]  x 2 ~ x 2 [HCN] (0.25-x) 0.25

13.  Example #6 cont….  K a = 6.2x10 -10 = [H 3 O + ][CN - ]  x 2 ~ x 2  [HCN] (0.25-x) 0.25 13 Fourth, solve for x which is equal to H 3 O + (6.2x10 -10 )(0.25) = x 2  x = 1.2x10 -5 M Finally, calculate the pH from the value of x, the [H 3 O + ] pH = -log(1.2x10 -5 ) = 4.90 *AP tips: Always follow these steps when performing calculations involving the Disassociation of weak acids or weak bases: 1.Write the reaction of the acid or base with water. 2.Set up an ICE chart 3.Write the equilibrium expression in terms of reactants and products. 4.Solve for x, using the method of approximation. Test approximation 5.Solve for pH. (be careful for base equilibria, x = OH-. You need to find pOH and then pH 6. Always remember that for weak acids and bases at eq: pH =-log[H3O+]

14.  The percent dissociation of an acid (or base) is the amount of the acid, HA, which has dissociated, x, divided by the acid’s initial concentration, HA 0, multiplied by 100  x/[HA 0 ] * 100%  When making assumptions in an equilibrium calculation, it is best to test the assumption by making sure that the percent dissociation is less than or equal to 5%.  The test for the assumption is the same as the calculation for the percent dissociation.  Example #7: The percent dissociation of an acid, HA, which is 0.100M is 2.5%. Calculate the K a of the acid. 14 x/0.100M x 100% = 2.5%  x = 2.5 x 10 -3 M K a = [H 3 O + ][A - ]  K a = (2.5x10 -3 ) 2 = 6.4x10 -5 [HA] (0.100 – 2.5x10 -3 )

15.  The calculations involving weak base equilibria are similar to the weak acid equilibria problems except that the equation is written for a base reacting with water and the calculation initially involves finding [OH-].  You will need to find the pOH and then the pH.  Example #8: The pH of a 0.20M solution of H 2 NNH 2 is 11.38. Calculate K b for H 2 NNH 2 15 Write the reaction with water. Bases accept H+. Watch charges! One hint to Help in writing the reaction is that the pH is 11.38. The basic pH indicates That OH- must be one of the products. H 2 NNH 2 + HOH   OH- + H 2 NNH 3 + I. 0.20 0 0 C. -x +x +x E. 0.20-x x x You are given the pH, but x=[OH-] Find pOH; pH+pOH=14.00 pOH= 14.00 – 11.38 = 2.62 pOH = -log[OH-] find [OH-] = inv log(-pOH) Inv log(-2.62) = 2.4x10 -3 M (you can type this Into your calculator as 10 -pOH Plug this value of x into the Kb expression: K b = [OH-][H 2 NNH 3 + ]/[H 2 NNH 2 ] = x 2 /(0.20-x) K b = (2.4x10 -3 ) 2 / 0.20 = 2.9x10 -5

16.  Polyprotic acids can donate more than one proton, H+, and dissociate by losing 1H+ at a time.  Example #9: Calculate the [H+] of a 0.20M solution.  Also determine the concentrations of H 3 AsO 4, H 2 AsO 4 -, HAsO 4 2-, and AsO 4 3-. 16 For H 3 AsO 4 ; K a1 = 5x10 -3, K a2 = 8x10 -8, K a3 = 6x10 -10 H 3 AsO 4 + HOH   H 3 O + + H 2 AsO 4 - I. 0.20 0 0 C. -x +x +x E. 0.20-x x x H 3 AsO 4 + H 3 O +   H 3 O + + H 2 AsO 4 - K a1 = [H 3 O + ][H 2 AsO 4 - ] / [H 3 AsO 4 ] 5x10 -3 = x 2 / (0.20-x) X = 3x10 -2 M [H 3 O + ]=[H 2 AsO 4 - ] = 3x10 -2 [H 3 AsO 4 - ] = 0.20 – 0.03 = 0.17M

17.  Example #9 cont…. For H 3 AsO 4 ; K a1 = 5x10 -3, K a2 = 8x10 -8, K a3 = 6x10 -10 Since K a3 <<<K a2 <<<K a1, very little of H 2 AsO 4 - and HAsO 4 2- dissociates compared to H 3 AsO 4, so [H 3 O + ] and [H 2 AsO 4 - ] will not change very much by the K a2 dissociation, and we can use their concentrations to find the concentration of HAsO 4 2-. H 2 AsO 4 - + HOH   H 3 O + + HAsO 4 2- I. 3x10 -2 3x10 -2 0 C. -x +x +x E. (3x10 -2 )-x (3x10 -2 )+x x K a2 = 8x10 -8 = (3x10 -2 )[HAsO 4 2- ] (3x10 -2 ) [HAsO 4 2- ] = 8x10 -8 M; the assumption that K a2 does not contribute significantly to [H 3 O + ] and [H 2 AsO 4 - ] is good. 17

18.  Example #9 cont…. For H 3 AsO 4 ; K a1 = 5x10 -3, K a2 = 8x10 -8, K a3 = 6x10 -10 Repeat the process to find [AsO 4 3- ] HAsO 4 2- + HOH   H 3 O + + AsO 4 3- I. 8x10 -8 3x10 -2 0 C. -x +x +x E. (8x10 -8 )-x (3x10 -2 )+x +x K a3 = 6x10 -10 = (3x10 -2 )[AsO 4 3- ] / (8x10 -8 ) [AsO 4 3- ] = 2x10 -15 M. Assumption that x is small is good. 18

19. Are both the cation and the anion neutral? Yes. The salt is neutral. No. Is one of the ions neutral? Yes, Is one of the ions acidic? Yes, salt is acidic. Cation is acidic, anion is neutral No, salt is basic. Cation is neutral, anion is basic. No One ion is acidic and one is basic. Ka>Kb acidic Ka<Kb it is basic Formula of salt split into ions Is cation a group I or II metal? Yes. Cation is neutral. Can’t accept or donate an H+ No. Cation is acidic. May be a c.acid of a weak base Is the anion a c.base of a strong acid? Yes. Anion is neutral. The c.base of a strong acid is neutral. No. Anion is basic. The c.base of a weak acid is basic. 19 Determining the Approximate pH Of ions in solution. Determining the Approximate pH Of a salt.

20.  Determine whether an aqueous solution of KC 2 H 3 O 2 is acidic, basic or neutral 20 KC 2 H 3 O 2 is basic. Use the method of asking questions outlined In the previous slide: Split the salt into its cation, K+, and its anion, C 2 H 3 O 2 - Is the cation of Group I metal? Yes; the cation does not affect the pH Is the anion a conjugate base of a strong acid? No; C 2 H 3 O 2 - is The c.base of a weak acid, HC 2 H 3 O 2, which makes the solution Basic. Because we have a salt with a cation which doesn’t affect the pH And a basic anion, an aqueous solution of the salt is basic.

21.  To calculate the pH of a salt, first you must decide whether the salt is acidic, basic, or neutral.  If the salt is basic, then its anion is the c.base of a weak acid  The anion will undergo hydrolysis.  You will need to write an equation for the reaction of that ion with water to form the acid and OH- ions: A- + HOH = HA + OH-.  Then write the equilibrium expression K b = K w = [HA][OH-] K a [A-]  If the salt is basic, then the hydrolysis reaction will produce an acid. 21

22.  Determine the pH of a 0.100M aqueous solution of NaCN. The K a for HCN is 5.8x10 -10 22 In aqueous solutions, NaCN ionizes completely into Na+ and CN-. However, the cyanide ions ionize in water according to the following: CN- (aq) + HOH (l)   HCN (aq) + OH- (aq) Na+ ions do not affect the pH. Ions from group IA and IIA never undergo Hydrolysis because they are cations of strong bases. The equilibrium Expression for the solution is: K b = [OH-][HCN] [CN-] We need a value for K b. Since we have K a for HCN, we can calculate the Value of K b for CN-: Kb = (1.0x10 -14 )/K a = (1.0x10 -14 )/(5.8x10 -10 ) CN- + HOH   HCN + OH- I. 0.100 0 0 C.-x +x +x E. 0.100-x x x Since K is much smaller than 0.100, Assume that 0.100-x = 0.100 K b = [OH-][HCN]/[CN-]  1.7x10 -5 = x 2 /0.100 X = [OH-] = 1.3x10 -3 pOH = 2.89; pH = 14.00 – 2.89 pH = 11.11

23.  Buffers resist changes in pH when acids or bases are added.  Calculating the pH of a buffer:  The calculation to find the pH of a buffer is similar to all equilibrium calculations EXCEPT there are now two initial concentrations, one for each part of the buffer pair. 23 ComponentsExamples Weak acid + salt containing the c.baseHCN & NaCN Weak base + salt containing the c.acidCH3NH2 & CH3NH3Cl Weak acid + excess strong base *or* weak base + excess strong acid 2 mol of HCN + 1mol NaOH react to yield 1mol HCN and 1mol NaCN 2mol NH3 + 1mol HCl react to yield 1mol NH3 and 1mol NH4Cl

24.  Calculate the pH of a solution that is 0.60M HF and 1.00M KF. K a for HF is 7.2x10 -4 24 First, write the reaction of the acid with water. (You are given K a, and The buffer is made of a weak acid and its c.base.) HF + HOH   H 3 O + + F - Second, set up an ICE chart I.0.60 0 0 C.-x +x +x E. 0.60-x x x Third, write the equilibrium Expression for K a. Plug in the values From the equilibrium line (E) of the ICE chart. Check to see if x is small. K a = 7.2x10 -4 = [H 3 O + ][F - ] = x(1.00-x) = x(1.00) [HF] (0.60-x) 0.60 Fourth, solve for x Which equals H 3 O + K a = 7.2x10 -4 = x(100)/0.60  x = [H 3 O + ] = 4.3x10 -4 pH = -log[H 3 O + ]  pH = -log(4.3x10 -4 ) = 3.37

25.  A buffer can be made from a weak acid and a salt containing its c.base or from a weak base and a salt containing its c.acid  Henderson-Hasselbalch equation: pH = pK a + log([c.base]/[acid])  Example #13: Calculate the mass of NaC 2 H 3 O 2 required to prepare a buffer of pH 4.55 when added to 0.500L of 0.67M acetic acid. (assume no change in volume) K a = 1.8x10 -5 for HC 2 H 3 O 2 25 Answer is 1.8g NaC 2 H 3 O 2 ……. This problem can be solved by either using an ICE chart or by using the Henderson-Hasselbalch equation. pK a = -logK a  pH = pK a + log([C 2 H 3 O 2 -] / [HC 2 H 3 O 2 ])  4.55 = 4.74 + log ([C 2 H 3 O 2 -] / 0.67M)  -0.19 = log([C 2 H 3 O 2 -]/0.67M)  0.65 = [C 2 H 3 O 2 -]/0.67M  0.44M = [C 2 H 3 O 2 -]  Mass NaC 2 H 3 O 3 = 0.44mol[C 3 H 3 O 2 -] x 1mol NaC 2 H 3 O 2 x 0.500L x 82.0g  1 L 1 mol [C 2 H 3 O 2 -] 1mol NaC 2 H 3 O 2

26.  You will solve these problems in the same manner as the previous problem #12, with the exception of the first three steps which are new.  Example #14: calculate the pH when 100.0 mL of 0.50M HCl are added to the buffer consisting of 20.0g HC 2 H 3 O 2 and 18.0g of NaC 2 H 3 O 2 dissolved in 5.00x10 2 mL of water. 26 First, Calculate the initial mols of weak acid and c.acid present in the buffer. Write These amounts under the reaction, above the ICE table. Second, calculate the moles of strong acid, H+, added. The added strong Acid will react with the basic part of the buffer, C 2 H 3 O 2 to produce the other Part of the buffer HC 2 H 3 O 2. Make a new line under the initial mole called ‘add mol H+’. Calculate the new initial moles of weak acid and it c.base After the c.base has reacted with the strong acid, by adding or subtracting the number of moles H+ added. HC 2 H 3 O 3 + HOH   H 3 O + + C 2 H 3 O 2 - Initial mols 0.33 0 0.21 Add mol H+ +0.05 0 -0.05 New initial mol 0.38 0 0.16 I. 0.63 0.27 C. -x +x +x E. 0.63-x x 0.27+x Initial [] of weak acid and c.base: HC 2 H 3 O 2 = 0.38mol/0.600L = 0.63M C 2 H 3 O 2 - = 0.16mol/0.600L = 0.27M Plug values into Henderson- Hasselbalch K a = [H+][C 2 H 3 O 2 ] [HC 2 H 3 O 2 ] 1.8x10 -5 =[H+](0.27+x)~[H+]0.27 (0.63+x) 0.63 4.2x10 -5 M=[H+] pH=-log[H+]  4.38

27.  Example #15: Calculate the pH when 100.0ml 0.50M NaOH are added to the buffer consisting of 20.0g of HC 2 H 3 O 2 and 18.0g of NaC 2 H 3 O 2 dissolved in 5.00x10 2 mL of water. 27 1 st, Calculate the initial moles of weak acid and c.acid that are present in the Buffer. Write these amounts under the reaction, above the ICE table. 2 nd, caclulate the mole of strong base, OH- added. The added strong base Will react with the acidic part of the buffer, HC 2 H 3 O 2 to produce the other Part of the buff, C 2 H 3 O 2 -. Make a new line under the initial mols called ‘add Mol OH-’. 3 rd, calculate the new initial concentration of the acid and its c.base and fill Out the ICE chart as in example #14. HC 2 H 3 O 2 + HOH   H 3 O + + C 2 H 3 O 2 - Initial mols 0.33 0 0.21 Add mol OH- -0.05 0 +0.05 New initial mol 0.28 0 0.26 I 0.47 0.43 C -x +x +x E 0.47-x x 0.43+x [H+] = 2.0x10 -5 M  pH = 4.71

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