H 2 O (l) H + (aq) + OH (aq)

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1 What is ph? Water not only serves as the solvent in solutions of acids and bases, it also plays a role in the formation of the ions. In aqueous solutions of acids and bases, water sometimes acts as an acid and sometimes as a base. You can think of the self ionization of water as an example of water assuming the role of an acid and a base in the same reaction: H 2 O H 2 O H 3 O + + OH equal Pure water contains concentrations of H + and OH ions produced by self ionization. One molecule of water acts as a Bronsted Lowry acid by donating a hydrogen ion/proton to the second water molecule, the Bronsted Lowry. base The 1:1 ratio between the products means that equal numbers of hydronium ions and hydroxide ions are formed. The equation for the equilibrium can be simplified in this way: H 2 O (l) H + (aq) + OH (aq) The double arrow indicates that this is in an. equilibrium Recall the equilibrium constant expression is written by placing the concentrations of the products over the concentration of the. reactants K eq = The concentration of pure water is constant so we write a new equilibrium constant, K w. K w = [H + ] ] (aq) [OH (aq) [ H 2 O (l) ] [H + (aq) ] ] [OH (aq) The constant, K w, is called the ion product constant for water: the value of the equilibrium constant expression for the self ionization of water. This is a special equilibrium constant that applies ONLY to the self ionization of water. Experiments show that in pure water at 298K, [H + ] and [OH ] are both equal to M. Therefore, at 298 K, the value of K w is: K w = [H + ][OH ] = ( M)( M) = 1.0 x No UNITS! 1

2 H 2 O (l) H + (aq) + OH (aq) ALWAYS The product of [H + ] and [OH ] equals at 298 K. This means that if the [H + ] increases, the [OH ] must. decrease Similarly, an increase in the [OH ] causes a in the [H + ]. decrease You can think about these changes in terms of Le Chatelier s principle: When a stress is applied to a system in equilibrium, the equilibrium will shift to relieve the stress. Adding extra hydrogen ions to the self ionization of water at equilibrium is a stress on the system. The system reacts in a way to relieve that stress. The extra hydrogen ions react with the hydroxide ions that are already present in the equilibrium to form more water molecules. Thus, the [OH ] decreases. The following example shows how you can use K w to calculate the concentration of either the hydrogen ion or the hydroxide ion if you know the concentration of the other ion. Example: At 298 K, the [H + ] of an aqueous solution is M. What is the [OH ] in the solution? Is the solution acidic, basic or neutral? Step 1: Analyze the problem Known Unknown [H + ]= 1.0x10 5 M [OH ]=? Kw = 1.0x10 7 Step 2: Solve for the unknown K w = [H + ] [OH ] [OH ] = K w [H + ] = 1.0 x x 10 5 = 1.0 x 10 9 M Step 3: Evaluate the Answer How do we know if the solution is acidic, basic or neutral? [H + ]= 1.0x10 5 M [OH ] = 1.0 x 10 9 M [H + ]>[OH ] Therefore, the solution is acidic. Question #1 2

3 ph and poh As you have probably noticed, concentrations of H + ions are often small numbers expressed in scientific notation. Because these numbers are somewhat hard to work with, chemists adopted an easier was to express [H + ] using a ph scale based on common logarithms. The ph of a solution is the negative logarithm of the hydrogen ion concentration: ph = log [H + ] below At 298 K, acidic solutions have ph values 7. Basic solutions have ph values above 7. Thus, a solution having a ph of 0.0 is a strong acid; a solution having a ph of 14.0 is a strong base. A solution with a ph of 7 is considered. neutral The logarithmic nature of the ph scale means that a change of one ph unit represents a tenfold change in ion concentration! Hence, a solution having a ph of 3.0 has 10 times the [H + ] of a solution with a ph of 4.0. The following figure shows the ph scale and ph values of some common substances: Example: Calculating ph from [H + ] What is the ph of a solution with a [H + ]= 1.0 x M concentration? ph = log [H + ] = log (1.0 x ) = NOTE: There are no units for ph. Values of ph are expressed with as many decimal places as the number of sig. figs. in the [H + ]. Because the concentration of [H + ] had 2 sig. figs, the ph should have 2 numbers after the decimal. Question #2 Using poh Sometimes chemists find it convenient to express the basicity, or, alkalinity of a solution on a poh scale that mirrors the relationship between ph and [H + ]. The poh of a solution is the negative logarithm of the hydroxide ion concentration: poh = log [OH ] 3

4 At 298 K, a solution having a poh less than 7.0 is ; basic a solution having a poh of 7.0 is neutral; and a solution having a poh greater than 7.0 is. acidic As with the ph scale, a change of one poh unit expresses a tenfold change in ion concentration. For example, a solution with a poh of 2.0 has 100 times the hydroxide ion concentration of a solution with a poh of 4.0. A simple relationship between ph and poh makes it easy to calculate either quantity if the other is known: ph + poh = The following figure illustrates the relationship between ph and [H + ] and the relationship between poh and [OH ] at 298 K. Use this diagram as a reference until you become thoroughly familiar with these relationships. Example: Calculating poh and ph from [OH ] An ordinary household ammonia cleaner is an aqueous solution of ammonia gas with a hydroxide ion concentration of M. Calculate the poh and the ph of a typical cleaner at 298 K. poh = log [OH ] = log (4.0 x 10 3 ) 2 sig. figs. = = #'s after the decimal! Solve for ph using the equation: ph + poh = ph = poh = = Question #3 4

5 Calculating ion concentrations from ph Suppose the ph of a solution is 3.50 and you must determine the concentrations of H + and OH. The formula for ph can be rearranged to solve for [H+]. ph = log [H + ] 1. Move the negative to the other side. ph = log [H + ] 2. To remove "log", use antilog (10 x button on calculator!) antilog ( ph) = [H + ] 3. Use your calculator to solve for [H + ] [H+] = antilog ( ph) [H+] = antilog ( ph) = antilog ( 3.50) = M (or 3.2 x 10 4 M) 2 #'s after the decimal = 2 sig. figs. with concentration! ph and poh don't have units but concentration does! You can now calculate [OH ] but there are 2 correct ways to do it! Option 1: You know [H + ], so use your Kw formula to solve for [OH ] Kw = [H + ][OH ] [OH ] = Kw [H + ] = 1.0 x x 10 4 M = 3.2 x M Option 2: You know ph, so solve for poh and then [OH ] ph + poh = poh = ph = = [OH ] = antilog ( poh) = antilog ( 10.50) = 3.2 x M concentration has units!!! Question #4 Calculating the ph of solutions of STRONG acids and bases We previously learned that strong acids and strong bases are essentially 100% ionized. Consider the following strong solutions: a) 0.1 M H b) 0.1 M NaOH Because H is a strong acid, that means that this reaction for the ionization of H goes to completion: H (aq) H + (aq) + (aq) 0.1 M 0.1 M 0.1 M Every H molecules produces one H + ion. The label 0.1 M H means that it contains 0.1 M mole H + ions per litre and 0.1 M per litre. For all strong monoprotic acids, the concentration of the acid IS the concentration of the H + ion. Thus, you can use the concentration of the acid as the [H + ] for calculating ph. 5

6 H H H H H 5 M H H HH H H 5M H and 5M Similarly, the 0.1 M solution of the strong base NaOH is also fully ionized: Na + (aq) + OH (aq) 0.1 M 0.1 M 0.1 M one NaOH (aq) One formula unit of NaOH produces OH ion. Therefore, since the concentration of NaOH is 0.1 M, the hydroxide ion concentration is. 0.1 M Some strong bases contain two or more hydroxide ions in each formula. Calcium hydroxide ( ) Ca(OH) 2 is an example. Because it has 2 hydroxide ions per molecule of calcium hydroxide, the concentration of the hydroxide ions is the concentration of the compound. twice OH Ca OH OH Ca OH OH Ca OH Ca Ca Ca OH OH OH OH OH OH 3M Ca(OH) 2 3M Ca and 6M OH Ex) What is the [OH ] in a M solution of calcium hydroxide? Ca(OH) 2(s) Ca 2+ (aq) + 2OH (aq) 7.5 x 10 4 M 7.5 x 10 4 M 2(7.5 x 10 4 M) = 1.5 x 10 3 M You can now use the concentration of the base for calculating poh and then ph. Question #5 Using ph to calculate K a Suppose you measured the ph of a M solution of the WEAK acid, HF, and found it to be Would you have enough information to calculate Ka for HF? YES!!! 6

7 You are given the initial concentration of the solution. You are also given the ph of the solution. From this, you can calculate the [H + (aq)] at equilibrium. Using this information and an ICE chart, you can then calculate the concentration of F. Once you know the equilibrium concentrations of all substances from your ICE chart, you can plug your values into your Ka formula. HF (aq) + H 2 O (l) H 3 O + (aq) + F (aq) [H 3 O + (aq)] [F (aq)] Ka = Ka = [HF (aq) ] HF (aq) H + (aq) + F (aq) [H + (aq)][f (aq)] [HF (aq) ] NOTE: [H 3 O + (aq)] and [H + (aq)] are equivalent to each other. If you know [H 3 O + (aq)], that IS your [H + (aq)]! I C E HF (aq) H + (aq)/h 3 O + (aq) F (aq) Use the ph to calculate [H + ] [H + ] = antilog ( ph) = antilog ( 2.38) = M Substitute known values into the Ka expression: Ka = [H+ (aq)][f (aq)] [HF (aq) ] = (0.0042)(0.0042) (0.096) = or 1.8 x 10 4 Question #6 7

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