Acids & Bases Strong & weak 1
Acid Base Dissociation Acid-base reactions are equilibrium processes. The relationship between the relative concentrations of the reactants and products is a constant for a given temperature. It is known as the Acid or Base Dissociation Constant (Ka or Kb). The stronger the acid or base, the larger the value of the dissociation constant Write the K for this equation: 2
Generally... 3
Acid Strength Acid strength depends on how much an acid dissociates (breaks up into ions). The more it dissociates (turns into H3O + ), the stronger it is The larger the Ka, the stronger the acid 4
Concentration vs. Strength Concentrations refers to how many moles are in a volume Strength refers to how much of the substance has ionized 5
Acid Strength Strong Acids - Transfers all of its protons to water; - Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated. - K is so large it is often not written Identify the conjugate acid-base pairs. Examples of Strong Acids Hydrochloric Acid Perchloric Acid Nitric Acid Sulfuric Acid 6
Weak acids Transfers only a fraction of its protons to water; - Partly ionized; - Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water 7
As acid strength decreases, base strength increases. The stronger the acid, the weaker its conjugate base The weaker the acid, the stronger its conjugate base 8
Base Strength Strong Base - all molecules accept a proton; Sodium Hydroxide - completely ionizes; Potassium Hydroxide - strong electrolyte; *Barium Hydroxide - conjugate acid is very weak, negligible *Calcium Hydroxide tendency to donate protons. Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons. As base strength decreases, acid strength increases. The stronger the base, the weaker its conjugate acid. The weaker the base the stronger its conjugate acid. 9
Try it! You want to know which is the stronger acid: HCl or CH3COOH. Here is your data. Write the equation for each acid in water Write the Ka for both acids Explain which is stronger, using your data 10
Try it! 11
Acids & Bases ph Scale 12
[H + ] ph 10-14 14 1 M NaOH 10-13 13 0 Acid 7 ph Scale Base The ph scale is a logarithmic scale used to assess acidity based on the equilibrium concentration of hydrogen or hydronium ion 14 Acidic Neutral Basic 10-12 12 10-11 11 10-10 10 10-9 9 10-8 8 10-7 7 10-6 6 10-5 5 10-4 4 10-3 3 10-2 2 10-1 1 10 0 0 Ammonia (household cleaner) Blood Pure water Milk Vinegar Lemon juice Stomach acid 1 M HCl
ph and acidity The ph values of several common substances are shown at the right. Many common foods are weak acids Some medicines and many household cleaners are bases. 14
ph of Common Substance ph [H 3 O + ] [OH - ] poh More basic More acidic NaOH, 0.1 M Household bleach Household ammonia Lime water Milk of magnesia Borax Baking soda Egg white, seawater Human blood, tears Milk Saliva Rain Black coffee Banana Tomatoes Wine Cola, vinegar Lemon juice Gastric juice 14 1 x 10-14 1 x 10-0 0 13 1 x 10-13 1 x 10-1 1 12 1 x 10-12 1 x 10-2 2 11 1 x 10-11 1 x 10-3 3 10 1 x 10-10 1 x 10-4 4 9 1 x 10-9 1 x 10-5 5 8 1 x 10-8 1 x 10-6 6 7 6 1 x 10 10-6 1 x 10 10-8 7 8 5 1 x 10-5 1 x 10-9 9 4 1 x 10-4 1 x 10-10 10 3 1 x 10-3 1 x 10-11 11 2 1 x 10-2 1 x 10-12 12 1 1 x 10-1 1 x 10-13 13 0 1 x 10 0 1 x 10-14 14
Calculating the ph The ph system is a logarithmic representation of the Hydrogen Ion concentration as a means of avoiding using large numbers and powers. ph = - log [H 3 O + ] Example 1: If [H 3 O + ] = 1 X 10-10 ph = - log 1 X 10-10 ph = 10 Example 2: If [H 3 O + ] = 1.8 X 10-5 ph = - log 1.8 X 10-5 ph = - (- 4.74) ph = 4.74 16
ph and Water Kw = [H 3 O + ] [OH - ] = 1.0 x10-14 In pure water [H 3 O + ] = 1 x 10-7 mol / L (at 20 o C) ph = - log(1 x 10-7) = - (0-7) = 7 ph < 7 (Acidic) [H 3 O + ] > 1 x 10-7 M ph > 7 (Basic) [H 3 O + ] < 1 x 10-7 M 17
poh and basicity The poh system is a logarithmic representation of the Hydroxide Ion concentration as a means of avoiding using large numbers and powers. poh = - log [OH - ] The relationship between ph and poh: ph + poh = 14 18
ph Calculations ph ph = -log[h 3 O + ] [H 3 O + ] = 10 -ph [H 3 O + ] ph + poh = 14 [H 3 O + ] [OH - ] = 1.0 x10-14 poh poh = -log[oh - ] [OH - ] = 10 -poh [OH - ]
ph = - log [H 3 O + ] Given: ph = 4.6 ph = - log [H 3 O + ] 4.6 = - log [H 3 O + ] - 4.6 = log [H 3 O + ] determine the [hydronium ion] choose proper equation substitute ph value in equation multiply both sides by -1 2 nd log 10 x antilog - 4.6 = log [H 3 O + ] take antilog of both sides [H + ] = 2.51x10-5 M Recall, [H 3 O + ] = [H + ] You can check your answer by working backwards. ph = - log [H + ] ph = - log [2.51x10-5 M] ph = 4.6
Indicators 21
Hydrangeas 22
Try it! 1. A certain solution has [H3O + ] = 2 x 10-5. Calculate the concentration of [OH - ] and the ph of the solution 2. A certain solution has [H3O + ] = 5 x 10-9. Calculate the concentration of [OH - ] and the ph of the solution 3. A certain solution has [OH - ] = 4 x 10-3. Calculate the concentration of [H3O + ] and the ph of the solution 4. A certain solution has [OH - ] = 3 x 10-11. Calculate the concentration of [H3O + ] and the ph of the solution 5. A certain solution has ph = 3.0. Calculate the poh, the concentration of [H3O + ] and the [OH - ] of the solution 6. A certain solution has ph = 2.5. Calculate the poh, the concentration of [H3O + ] and the [OH - ] of the solution 7. A certain solution has poh = 2.0. Calculate the ph, the concentration of [H3O + ] and the [OH - ] of the solution 8. A certain solution has poh = 9.3. Calculate the ph, the concentration of [H3O + ] and the [OH - ] of the solution 23
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Neutralization An acid will neutralize a base, giving a salt and water as products Examples HCl + NaOH NaCl + H 2 O H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O H 3 PO 4 + 3 KOH K 3 PO 4 + 3 H 2 O 2 HCl + Ca(OH) 2 CaCl 2 + 2 H 2 O 25
Neutralization Calculations Recall: n = v x c Where: n represents the number of moles in mol v represents volume in dm 3 c represents concentration in mol/ dm 3 or mol dm -3 26
Neutralization Problems If an acid and a base combine in a 1 to 1 ratio, the moles of acid will equal the moles of base nacid = nbase Therefore the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base V acid C acid = V base C base If any three of the variables are known it is possible to determine the fourth 27
Neutralization Problems Example 1: Hydrochloric acid reacts with potassium hydroxide according to the following reaction: HCl + KOH KCl + H 2 O If 15.00 cm 3 of 0.500 M HCl exactly neutralizes 24.00 cm 3 of KOH solution, what is the concentration of the KOH solution? Solution: V acid C acid = V base C base (15.00 cm 3 )(0.500 M) = (24.00 cm 3 ) C base C base = (15.00 cm 3 )(0.500 M) (24.00 cm 3 ) C base = 0.313 M 28
Neutralization Problems Whenever an acid and a base do not combine in a 1 to 1 ratio, a mole factor must be added to the neutralization equation n V acid C acid = V base C base The mole factor (n) is the number of times the moles the acid side of the above equation must be multiplied so as to equal the base side. (or vice versa) Example H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O The mole factor is 2 and goes on the acid side of the equation. The number of moles of H 2 SO 4 is one half that of NaOH. Therefore the moles of H 2 SO 4 are multiplied by 2 to equal the moles of NaOH. 29
Neutralization Problems Example 2: Sulfuric acid reacts with sodium hydroxide according to the following reaction: H 2 SO 4 + 2 NaOH Na 2 SO 4 + 2 H 2 O If 20.00 cm 3 of 0.400 M H 2 SO 4 exactly neutralizes 32.00 cm 3 of NaOH solution, what is the concentration of the NaOH solution? Solution: In this case the mole factor is 2 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore 2 V acid C acid = V base C base 2 (20.00 cm 3 )(0.400 M) = (32.00 cm 3 ) C base C base = (2) (20.00 cm 3 )(0.400 M) C base (32.00 cm 3 ) = 0.500 M 30
Neutralization Problems Example 3: Phosphoric acid reacts with potassium hydroxide according to the following reaction: H 3 PO 4 + 3 KOH K 3 PO 4 + 3 H 2 O If 30.00 cm 3 of 0.300 M KOH exactly neutralizes 15.00 cm 3 of H 3 PO 4 solution, what is the concentration of the H 3 PO 4 solution? Solution: In this case the mole factor is 3 and it goes on the acid side, since the mole ratio of acid to base is 1 to 3. Therefore 3 V acid C acid = V base C base 3 (15.00 cm 3 )(C acid ) = (30.00 cm 3 ) (0.300 M) C acid = (30.00 cm 3 )(0.300 M) C acid (3) (15.00 cm 3 ) = 0.200 M 31
Neutralization Problems Example 4: Hydrochloric acid reacts with calcium hydroxide according to the following reaction: 2 HCl + Ca(OH) 2 CaCl 2 + 2 H 2 O If 25.00 cm 3 of 0.400 M HCl exactly neutralizes 20.00 cm 3 of Ca (OH) 2 solution, what is the concentration of the Ca(OH) 2 solution? Solution: In this case the mole factor is 2 and it goes on the base side, since the mole ratio of acid to base is 2 to 1. Therefore V acid C acid = 2 V base C base (25.00 cm 3 ) (0.400) = (2) (20.00 cm 3 ) (C base ) C base = (25.00 cm 3 ) (0.400 M) C base (2) (20.00 cm 3 ) = 0.250 M 32