Unit 5- Acids and bases 5.2.1- Strong acids and bases 5.2.2- Ionization constants 5.2.3- Calculating pH 5.2.4- The pH scale 5.2.5- Indicators Unit 5- Acids and bases

5.2.1 Strong & Weak Acids & Bases Strong acids produce many H+ ions (or H3O+ ions) weak acids produce few H+ ions Strong bases produce many OH- ions weak bases produce few OH- ions

Strong Acids Strong acids and bases are essentially one-way reactions - the acid or base breaks down completely to produce ions. At equilibrium there are only products (the ions) left. Strong Acids HCl(aq) → H+(aq) + Cl-(aq) H2SO4 (aq) → H+(aq) + HSO41-(aq) Strong Bases NaOH(aq) →Na+(aq) + OH- (aq) Mg(OH)2 (aq) → Mg2+(aq) + 2 OH-(aq)

Weak Acids Weak acids and bases do not ionize completely. For weak electrolytes, equilibrium lies to the reactant side of the equation where there will be few ions present. Some examples: Weak Acids HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq) HCHO2 (aq) ↔ H+(aq) + CHO2-(aq) Weak Bases NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq) NH2CH3(aq) + H2O(l) ↔ NH3CH3+(aq) + OH-(aq)

5.2.2 Ionization Constants: Ka, Kb, and Kw Ka and Kb- Pay attention to the physical states and whether or not a particular substance is included in the equilibrium Note: Ka for acids and Kb for bases.

Ka values HCl(aq) → H+(aq) + Cl-(aq) Ka = [H+] [Cl-] =1.3×106 [HCl] HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)  Ka = [H3O+] [Cl-] =1.3×106 HCHO2 (aq) ↔ H+(aq) + CHO2-(aq)  Ka = [H+] [CHO2-] = 1.8 × 10-4 [HCHO2]

Kb values Mg(OH)2 (aq) → Mg2+(aq) + 2 OH-(aq) Kb = [Mg2+] [OH-]2 NH3 (aq) + H2O(l) ↔ NH4+(aq) + OH-(aq)  Kb = [NH4+] [OH-] = 1.8×10-5 [NH3]

A large value of Ka  means there are many H+ ions in solution - in other words, a strong acid A large Kb indicates many OH- ions - a strong base The Table of Acid and Base Strengths gives Ka and Kb values for a number of acids and bases. Kavalues for very strong acids are not given

Kw We usually do not think of water as producing ions, but water does ionize, but not very well. We can write the ionization equation for water in two ways. The equilibrium constant, Kw, can also be written for both equations Notice that H2O does not appear in the Kw expression because it is a liquid H2O(l) → H+(aq) + OH-(aq)  Kw = [H+] [OH-] = 1.0 × 10-14  2 H2O(l) → H3O+(aq) + OH-(aq)  Kw = [H3O+] [OH-] = 1.0 ×10-14 Values for Kw are given for 25°C.

Two key items to note: In pure water, the balanced equation tells us that the concentrations of H+ and OH- will be equal to one another. We find that [H+] = [OH-] = 1.0×10-7 The value of Kw is very small, meaning that very few ions are present. As long as temperature remains constant Kw is a constant and it's value will not change.

5.2.3 Calculating [H+] and [OH–] Ka and Kb are used to calculate the concentrations of ions in acids and bases. This will be crucial to determining pH How you calculate ion concentrations depends on whether you have a strong acid (or base) or a weak acid (or base).

Calculating Ion Concentrations for Strong Acids & Bases For strong acids and bases, the concentration of the ions can be readily calculated from the balanced equation.

Examples 1.Calculate the hydrogen ion concentration in a 0.050 M solution of hydrochloric acid. Solution: We know that HCl is a strong acid that ionizes completely in water (you should memorize the list of strong acids). Begin by writing the balanced reaction: HCl(aq) → H+(aq) + Cl-(aq) From the balanced equation we see that 1 mole of HCl produces 1 mole of H+ (a 1:1 ratio), therefore the concentration of H+ will equal that of HCl. Answer: [H+ ] = 0.050 M Also [Cl-] = 0.050 M

Example #2 2.Calculate the hydroxide ion concentration in a 0.010 M solution of barium hydroxide, Ba(OH)2. Barium hydroxide is a strong base. 

Solution: Always begin by writing a balanced equation: Ba(OH)2 (aq) → Ba2+(aq) + 2 OH-(aq) Since 2 moles of OH- are produced for every 1 mole of Ba(OH)2 , the concentration of OH- will be twice the concentration of Ba(OH)2 . Answer: [OH-] = 2 × 0.010 = 0.020 M Also [Ba2+] = 0.010 M

Calculating Ion Concentrations for Weak Acids & Bases Weak acids and bases require a much different approach to finding ion concentrations. Once you know you have a weak acid or base, follow these steps: Write a balanced equation You will need to know the value of Ka or Kb, look it up in a Table of Acid and Base Strengths. Set up the equilibrium constant expression.

EXAMPLE 1.Calculate the hydrogen ion concentration in a 0.10 M acetic acid solution, C2H3O2.  Ka for acetic acid, a weak acid, is 1.8 ×10-5. 

Solution: Begin by writing the balanced reaction: HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq) Concentration of the acid, HC2H3O2 is 0.10 M. We need to find the concentration of H+, which will also equal the concentration of C2H3O2- Because ionization is NOT complete because this is a weak acid, [H+] will NOT equal [HC2H3O2]. Instead we must calculate it using the equilibrium constant expression.

Set up the Ka equation: Ka = [H+] [ C2H3O2-] [HC2H3O2] Substitute values into the equation. Let x equal the unknowns 1.8 ×10-5 = (x) (x) 0.10 Rearrange the equation X2 = (1.8 ×10-5)(0.10) X2 = 1.8 ×10-6 Take the square root X = 1.3×10-3 ANSWER: [H+] = 1.3×10-3 Also [C2H3O2-] = 1.3×10-3

Example 2.Calculate the hydroxide ion concentration, [OH-], in a 0.025 M solution of analine, C6H5NH2, a weak base with Kb = 4.3×10-10 

Solution: Begin by writing a balanced equation. Since analine is a base that doesn't contain the hydroxide ion, include H2O as a reactant. C6H5NH2 (aq) + H2O (l) ↔ C6H5NH3+(aq) + OH-(aq) Set up the Kb expression and solve for ion concentrations. We see from the balanced equation that the ions have a 1:1 ratio, therefore [OH-] will equal the [C6H5NH3+]. Set up the Kb equation, omitting liquid water: Kb = [C6H5NH3+] [OH- ] [C6H5NH2]    Substitute values into the equation. Let x equal the unknowns 4.3×10-10 = (x) (x) 0.025 X2 = (4.3×10-10)(0.25)  X2 = 1.1×10-11 Take the square root X = 3.3×10-6 ANSWER:[OH-] = 3.3×10-6 M Also [C6H5NH3+] = 3.3×10-6 M

Finding [OH-] in Weak Acids and [H+] in Weak Bases H2O(l) → H+(aq) + OH-(aq); [H+= 0.050 M] Kw = [H+] [OH-] = 1.0 × 10-14 remember that equilibrium constants are constant. Thus the value of Kw will still have a value of 1.0 ×10-14 even if [H+] has increased due to the presence of the acid. We can use this information to calculate the concentration of hydroxide ions present in the aqueous solution:  Kw = [H+] [OH-] Rearrange the equation [OH- ] = Kw______ [H+] Substitute known values and solve for [OH- ] [OH- ] = 1.0 ×10-14 0.05 [OH- ] = 2.0 ×10-13

For any acid or base you can calculate  both [H+] and [OH-] Acids First determine [H+] then use Kw to calculate [OH-] Bases First determine [OH-] then use Kw to calculate [H+]  

Le Châtalier's Principle Recall: When we disrupt an equilibrium system by increasing the concentration of a reaction participant, equilibrium will shift to minimize the stress. When we increase the H+ ion concentration in the water equilibrium, the reaction will shift to the left to "use up" the additional H+. This will cause the concentration of OH- to decrease. Indeed, we see that [OH-] does decrease, from 1.0×10-7 in pure water to 2.0 ×10-13 in our acid solution.

Practice set 5.2.4- do questions 1-7

5.2.4 The pH Scale pH is just another way to express the hydrogen ion concentration ([H+]), of an acidic or basic solution.

What about pH? pH is defined as the negative log of hydrogen ion concentration. Mathematically it looks like this: pH = - log [H+]

Try these on your calculator [H+] pH 1 ×10-3 3.0 2.5 ×10-11 10.6 4.7 ×10-9 8.3 5.8 ×10-4 3.2 1.0×10-7 7.0

pH values Acids Bases Neutral solutions pH < 7 pH > 7 pH = 7 The lower the pH, the stronger the acid The higher the pH, the stronger the base

Examples of Calculating pH 1.Calculate the pH of a 0.01M HNO3 solution? 

Solution: Begin finding pH by first finding [H+]. HNO3 is a strong acid,based on the balanced equation, we see that there is a 1:1 ratio between HNO3 and H+, so [HNO3] = [H+]: HNO3 (aq) → H+(aq) + NO3-(aq) [H+] = -log [H+]  = -log (0.01)   = - (-2.0)  = 2.0 answer

Example #2 Find the pH of a 0.01 M solution of ammonia. Ammonia is a weak base with Kb= 1.8 × 10-5

Solution: This is a more difficult question because it’s a weak base (similar for a weak acid). We must use Kb to determine ion concentrations. That will require a balanced equation- for weak bases (that don't contain an OH–), be sure to include water, H2O as a reactant. Remember that the base will gain a hydrogen ion: NH3 (g) + H2O(l) ↔ NH4+(aq) + OH-(aq)

First calculate [OH-]. Then use Kw to determine [H+] Set up the Kb equation: Kb = [NH4+] [OH- ] [NH3] Substitute values into the equation. Let x equal the unknowns 1.8 ×10-5 = (x) (x) 0.01 Rearrange the equation X2 = (1.8 × 10-5) (0.01) X2 = 1.8 × 10-7 Take the square root X = 4.2 × 10-4 [OH-] = 4.2 × 10-4 M

Next we calculate [H+]:  Kw = [H+] [OH-] [H+ ] = __Kw___ [OH-] Substitute in known values and calculate [H+] [H+] = 1.0 ×10-14 4.2 × 10-4 [H+] = 2.4 × 10-11 Finally, convert [H+] into pH: pH= -log[H+]  = -log(2.4 × 10-11)  pH= 10.6 answer 

pOH  There is a way to simplify the last parts of this operation, by using pOH: pOH = - log [OH-] Once we find [OH-] for a base, we can determine pOH: [OH-] =4.2 × 10-4 pOH= -log [OH-]  = -log (4.2×10-4)  pOH= 3.4 answer Next we use of the following easy-to-memorize relationship: pH + pOH = 14 Once we find pOH, it is a simple matter to find pH: OR pH = 14 - pOH   = 14 - 3.4  pH= 10.6

It doesn't matter which method you use to find [H+] and pH for a base - both will give you the same answer. Choose whichever method works best for you.

Do questions 8,9,11 from the practice problems 4-2-4

Finding [H+] when you know pH You know the pH of a solution and need to find [H+], or the concentration of the acid solution. How do you do that? To convert pH into [H+] involves taking the antilog of the negative value of pH .[H+] = antilog (-pH) different calculators work slightly differently - make sure you can do the following calculations using your calculator. Practice as we go along . . .

Example. We have a solution with a pH = 8.3. What is [H+]? Enter 8.3 as a negative number Use your calculator's 2nd or Shift or INV key to type in the symbol found above the LOG key. The shifted function should be 10x. You should get the answer 5.0 × 10-9

Other calculators require you to enter keys in the order they appear in the equation. Use the Shift or second function to key in the 10x function. Use the +/- key to type in a negative number, then type in 8.3 You should get the answer 5.0 × 10-9 If neither of these methods work, try rearranging the order in which you type in the keys.

Example 1.Find the hydronium ion concentration in a solution with a pH of 12.6. Is this solution an acid or a base? How do you know? 

Solution: The solution is a base because pH > 7. To find hydronium ion concentration, [H3O+], which will be the same as [H+]. To convert pH into [H3O+]: [H3O+] = antilog (-pH)  = antilog (-12.6)  [H3O+] = 2.5 × 10-13 answer 

Solution: In order to determine Ka  we need to know the concentrations of several things: [H+], [HCO3-], and [H2CO3] Begin with a balanced equation for the acid: H2CO3 (aq) ↔ H+(aq) + HCO3-(aq) Next set up the equilibrium expression, which will be needed to find Ka: Ka = [H+][HCO3-] [H2CO3]  Next find [H+] from pH.[H+] = antilog (-pH)  = antilog (-3.49)  [H+] = 3.2 × 10-4 The balanced equation tells us that [H+] = [HCO3-]. Substitute values into our Ka expression and solve: The question gave us the concentration of the acid, H2CO3, as 0.24 M  =[3.2 × 10-4] [3.2 × 10-4] [0.24]    Ka=4.3 ×10-7 answer

Example #2 A 0.24M solution of the weak acid, H2CO3, has a pH of 3.49. Determine Ka for H2CO3 (carbonic acid). 

Assignment 5.2.4

5.2.5 Indicators Indicators are dyes that change colour under varying conditions of acidity. Litmus is an indictor that changes colour from red to blue in the pH range of 5.5 to 8.0. Other indicators and their colours are listed in the table of Acid - Base Indicators

Indicators may be in solution form or paper form Indicators may be in solution form or paper form. pH paper is prepared by treating the paper with the indictor solution. When the paper is then dipped into the solution you are testing, it will change colour depending on the acidity of the solution. Here are some questions to try. A short acid-base indicator table is given here: Indicator pH range Colour change methyl orange 3.2 - 4.4 red to yellow Litmus 5.8 - 8.0 red to blue Phenolphthalein 8.2 - 10.0 colourless to pink

Example A given solution turns methyl orange yellow, litmus blue, and phenolphthalein red. What is the approximate pH of the solution?

Solution: Methyl orange in yellow when pH is above 4.4 Litmus is blue when pH is above 8.0, and Phenolphthalein is red when pH is above 10.0. Therefore the solution would have to have a pH above 10.0

a. vinegar (pH = 3) b. sea water (pH = 8) What color would methyl orange, litmus, and phenolphthalein turn when testing: a. vinegar (pH = 3) b. sea water (pH = 8) methyl orange litmus phenolphthalein vinegar red colourless sea water yellow red-blue