# The Concept of Chemical Equilibrium

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1 Unit 2 Strong Acids and Bases Chem 020, R. R. Martin The Concept of Chemical Equilibrium A chemical reaction is at equilibrium when it is proceeding in the forward and backward directions at the same rate. The concentrations of all the species remain constant once equilibrium is established. For the reaction: a A + b B cc + d D at equilibrium, the equilibrium constant is defined as: K eq = [C] c [D] d / [A] a [B] b Where [A] represents the concentration of reactant A at equilibrium, [B] the concentration of B, etc.. K eq is the Equilibrium Constant for the reaction at that temperature. Many important reactions can be considered as going to completion: HCl (g) + H 2 O (l) H 3 O + (aq) + Cl - (aq) or HCl (g) + H 2 O (l) H + (aq) + Cl - (aq) 1

2 How do we most often measure concentrations? Molarity: M Concentration is usually expressed as the number of moles of solute per litre of solution, mol L 1. M = (Moles solute)/(volume of solution in liters) Thus a 0.10 molar solution has a concentration of 0.10 mol L -1, and is written as 0.10 M. Generally square brackets mean moles per liter ie [A] Molarites can be used to calculate: The number of moles of solute in a given volume of solution or The volume of solution containing a given # of moles of solute. These calculations permit us to use volumetric glassware, burette, pipette, volumetric flask (Lab. Manual) to prepare and measure out chemical reagents. Examples 1. What is the molarity of a solution containing 1.50 mol of solute in 750 ml of solution? M = Moles/Volume (in litres) = Moles/ V = 1.5/.75 = 2.0 Therefore: Moles = MV = 8 x.05 = 0.4 2

3 2. How many moles of HBr are there in 50.0 ml of 8.00 M solution? Molarity = Moles/ V Moles = Molarity x Volume = 8 moles/ liter x.050 liters = 0.4 moles HBr 3. What volume (ml) of M HBr solution contains 2.40 mol of HBr? M = Moles / V moles L 1 = 2.40 moles/ V V = 0.2 L = 200 ml 3

4 Strong Aids and Bases An acid is defined as a substance that increases the Hydrogen Ion concentration when dissolved in water. The Hydrogen ion, H +, is a bare proton. It is never found as such in solution or in an ionic compound The ion H 3 O + is the hydronium ion. All H + in water is in this form. We write H 3 O +, H + (aq) or (lazily) just H + Some examples of acids in water HCl + H 2 O H + (aq) + Cl(aq) HNO 3 + H 2 O H + (aq) + NO 3 For strong acids we assume that reaction goes to completion, i.e. these acids are 100% dissociated (or Ionized). These contain hydrogen, which dissociates as a proton (or more properly a hydronium ion in solution and hence are called protonic acids. HCl, HBr and HNO 3 are monoprotic, giving 1 H + per molecule; i.e., 1 mol acid 1 mol H + (aq) H 2 SO 4 and H 2 CO 3 are diprotic, giving 2 H + per molecule and H 3 PO 4 is triprotic. E.g., H 2 SO 4 + H 2 O 2 H + (aq) + SO 4 2 (The discussion of weak acids, which do not dissociate completely, will be left to later in the course.) 4

5 In many cases different names are used for the pure substance and for the aqueous solution, those generally considered to be strong acids (ionize completely in water) are underlined HCl HBr HI HNO 3 HNO 2 H 2 SO 4 H 2 SO 3 hydrochloric acid hydrobromic acid hydroiodic acid nitric acid nitrous acid sulfuric acid sulfurous acid 5

6 Bases are substances which either 1. Increase OH ion concentration in water, or 2. Decrease H 3 O + ion concentration in water To see why these statements are equivalent, we must consider the autoionization of water again. It is an equilibrium 2 H 2 O H 3 O + (aq) + OH (aq) and K eq = [H 3 O+] [OH] / [H 2 O] 2 where [ ] means the concentration of the species within the braces in mol L 1 But in aqueous solution, the concentration of water, [H 2 O], is very nearly constant and = 55 mol L 1 (55 molar). (1 L has a mass of 1000 g and contains 1000/MM H 2 O mols of water = 1000 g L 1 /18 g mol 1 = 55 mol L 1 ) Omitting [H 2 O] from the K equ expression and defining the new reduced equilibrium constant as K w gives K w = [H 3 O + ] [OH ] = 1.0 x mol 2 L 2 or more simply as [H + ] [OH ] this is called the Ion Product of Water Its numerical value at 25º is 1.0 x mol 2 L 2 Since [H3O] + = [OH] in pure water then it follows [H 3 O + ] = [OH ] = (1 x ) = 1 x 10 7 mol L 1 = 1 x 10 7 M This is a very small concentration of ions. Pure water is not good conductor of electricity. 6

7 Note that [H 3 O] + and [OH ] are inversely related as [H 3 O + ] increases, [OH ] decreases. Thus in acidic solution as [H 3 O + ] is high, [OH ] must be very low. For example: In 0.1 M HCl, the acid is 100% dissociated (remember; it is a strong acid) and hence the [H 3 O + ] concentration is also 0.1 M K w = [H 3 O + ] [OH ] = 1 x M 2 [0.1] [OH ] = 1 x M 2 [OH ] = 1 x M 7

8 Examples of Bases: 1. Soluble hydroxides of the Group 1 metals NaOH + H 2 O Na + (aq) + OH (aq) 2. NH 3 + H 2 O NH + 4 (aq) + OH (aq) ammonia ammonium In 0.1 M NaOH solution, [OH ] = 0.1 M [H + ] = K w / [OH ] = (1.0 x )/0.1 = 1.0 x M Many metal oxides and hydroxides are insoluble in water, but act as bases by reacting with acids by decreasing [H +] Mg(OH) 2 (s) + 2 H + Mg H 2 O ZnO(s) + 2 H + Zn 2+ + H 2 O Mg(OH) 2 is acting as a base. As does zinc oxide 8

9 The reaction of acid and base (Neutralization) occurs very rapidly. The key reaction is: Example: H + + OH H 2 O HBr + NaOH NaBr + H 2 O Strong acid strong base ionic salt H + + Br Na + + OH Na + + Br The Na + and Br are spectator ions, taking no part in the reaction except to balance charge. 9

10 Expressing Acidity the ph Scale The quantity ph (power of hydrogen) is defined as ph = log [H + ] The ph scale is a convenient one for expressing the concentration of H + (hydronium) ions. In pure water, [H + ] = 1.0 x 10 7 ph = log ( 1.0 x 10 7 ) = 7.00 This is called a neutral solution. Because of the minus sign, ph decreases as [H + ] in solution increases ph < 7.00 [H + ] > 1.0 x 10 7 solution is acidic ph > 7.00 [H + ] < 1.0 x 10 7 solution is basic Now some example problems: 1. A solution has [H + ] = 1.6 x 10 3 M. What is the ph? Log (1.6 x 10 3 ) = 2.80 ph = (acidic) 2. What is [H + ] in a solution of ph 8.63? Remember that log[h + ] = 8.63, and that therefore Log [H + ] = 8.63 [H + ] = = 2.3 x 10 9 M (basic) [H + ] = 10 ph Calculate [OH ] in each of the above cases Remember that [H + ][OH] = 1.0 x = K w 1. [OH ] = K w / [H + ] = 1.0 x / 1.6 x 10 3 = 6.3 x M 2. [OH ] = 1.0 x / 2.3 x 10 9 = 4.3 x 10 6 M 10

11 Similarly, we use poh to express [OH ] concentration poh = log [OH ] Since [H + ][OH ] = K w Take logs to get Log [H + ] + log[oh ] = logk w or if pk w = log K w ph + poh = pk w = (at 25º) or poh = 14 ph As ph ranges from 0 (acidic) to 14 (basic), poh ranges from 14 to 0. e.g., in example 1 above, ph = 2.80 poh = = Stoichiometry of reaction for acids and bases is easy H + and OH react in 1:1 ratio Concentrations are always given in molarities, M units of mol L 1. And. Acid + base = salt + water 11

12 Problems are of two types 1. Neutralization equal amounts of H+ and OH 2. Mixing problems either H+ or OH EXAMPLES What volume of M NaOH will neutralize 25.0 ml of M HCl? The reaction is: HCl + NaOH NaCl + H 2 O 1 mol of HCl reacts with 1 mol NaOH Now from the definition of molarity we know: that for neutralization: Moles of Acid = Moles of Base And M (acid) x V (acid in litres) = Moles of Acid = Moles of Base = M (base) x V (base in litres) Or M A V A = M B V B x = x V B V B =.033 L = 33 ml 12

13 Another EXAMPLE 5.00 g NaOH is dissolved in 750 ml of water. A 25.0 ml portion neutralizes 16.0 ml of H 2 SO 4 solution. What is the H 2 SO 4 concentration? Write the balanced reaction H 2 SO4 + 2 NaOH Na 2 SO H 2 O Note that 2 moles of NaOH react with one mole of H 2 SO 4. If we know how many moles of NaOH reacted we can easily calculate the number of moles H 2 SO 4 present and hence the concentration. Begin with concentration of NaOH: M NaOH Moles NaOH/Volume solution = (mass NaOH/Molar mass NaOH)/Volume M NaOH = (5/40)/0.75 = Now the number of moles of NaOH = MV = x = moles Therefore the number of moles H 2 SO 4 = 4.17 x 10 3 /2 = 2.08 x 10 3 And M A V A = moles acid = 2.08 x 10 3 = MA x So that: M A =

14 Yet another EXAMPLE An organic acid is known to be diprotic. When g is dissolved in water, it requires 17.5 ml of M KOH to neutralize. What is the molar mass of the acid? Since we are told the acid is diprotic, we know that each mol of the acid will require 2 mol of KOH First, compute the number of mol of KOH = M KOH V KOH in litres = (17.5 ml / 1000 ML L 1 ) x = x 10 3 mol This will react with x 10 3 /2 = 7.81 x 10 4 mol acid So now we now know that: g of this acid is 7.81 x 10 4 mol MM = # g / # mol = g / 7.81 x 10 4 mol = 160 g mol 1 14

15 A Special Class of Problems: MIXING PROBLEMS When acids and bases are mixed, an excess of one often remains (Either the acid or the base turns out to be the limiting reagent). For example, What is the ph of a solution made by mixing 25.0 ml of M NaOH with 16.0 ml of M HCl? (As always write down the reaction: HCl + NaOH NaCl + H 2 O) Compute # of mols: NaOH: 25.0 / 1000 x = mol OH HCl: 16.0 / 1000 x = mol H + There is therefore an excess of OH : mol = mol But the total volume of solution is = 41.0 ml = L [OH ] = mol / L = M poh = log[0.0268] 1.57; ph = =

16 Another Mixing EXAMPLE What mass of solid Ca(OH) 2 must be added to 10.0 L of an acidic solution to change its ph from 1.0 to 2.5? Since of solid Ca(OH) 2 is used, assume no change in solution volume. We are given 2 values of ph, so we can compute the [H +] values in each case For the ph = 1 sol n : [H + ] = 10 1 M = M 10 L contains 10 L x M = 1.00 mol H + At ph = 2.5, [H + ] = = 3.2 x 10 3 M So 10 L contains 10.0 x 3.2 x 10 3 mol = mol H + Adding Ca(OH) 2 therefore decreased [H + ] by mol H+ (= mols of OH added) (Remembering that: Ca(OH) H + Ca H 2 O) Need 0.97 mol /2 of Ca(OH) 2 = 0.48 mol = 0.48 x MM g of Ca(OH) 2 = 0.48 x 74 g Ca(OH) 2 = 36 g Ca(OH) 2 16

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