[H ] [OH ] 5.6 " 10

Transcription

1 Howemork set solutions 10: 11.1 Table 11.5 of the tet contains a list of important Brønsted acids and bases. (a) both, base, (c) acid, (d) base, (e) acid, (f) base, (g) base, (h) base, (i) acid, (j) acid In general the components of the conjugate acid base pair are on opposite sides of the reaction arrow. The base always has one fewer proton than the acid. (a) (c) (d) (e) The conjugate acid base pairs are (1) HCN (acid) and CN (base) and () CH 3 COO (base) and CH 3 COOH (acid). (1) HCO 3 (acid) and CO 3 (base) and () HCO 3 (base) and H CO 3 (acid). (1) H PO 4 (acid) and HPO 4 (base) and () NH 3 (base) and NH 4 (acid). (1) HClO (acid) and ClO (base) and () CH 3 NH (base) and CH 3 NH 3 (acid). (1) H O (acid) and OH (base) and () CO 3 (base) and HCO 3 (acid) (a) Lewis acid; see the reaction with water shown in Section 11.1 of the tet. Lewis base; water can donate a pair of electrons to H to form H 3 O. (c) Lewis base. (d) Lewis acid; SO reacts with water to form H SO 3. Compare to CO above. Actually, SO can also act as a Lewis base under some circumstances. (e) (f) Lewis base; see the reaction with H to form ammonium ion. Lewis base; see the reaction with H to form water. (g) Lewis acid; does H have any electron pairs to donate? (h) Lewis acid; compare to the eample of NH 3 reacting with BF (a) Ba(OH) is ionic and fully ionized in water. The concentration of the hydroide ion is M (Why? What is the concentration of Ba?) Assuming activities can be approimated by concentrations, we find the hydrogen ion concentration.! 14 Kw 1.0 " 10! 11 = = = "! [H ] [OH ] 5.6 " 10 The ph is then: ph log[h ] = log( ) = M Nitric acid is a strong acid, so the concentration of hydrogen ion is also M. The ph is: ph = log[h ] = log( ) = 3.8 ph = log[h ] = log[ ] = (a) strong acid, weak acid, (c) strong acid (first stage of ionization), (d) weak acid, (e) weak acid, (f) weak acid, (g) strong acid, (h) weak acid, (i) weak acid.

2 11.40 Strategy: Recall that a weak acid only partially ionizes in water. We are given the initial quantity of a weak acid (CH 3 COOH) and asked to calculate the concentrations of H, CH 3 COO, and CH 3 COOH at equilibrium. First, we need to calculate the initial concentration of CH 3 COOH. In determining the H concentration, we ignore the ionization of H O as a source of H, so the major source of H ions is the acid. We follow the procedure outlined in Section 11.4 of the tet. Solution: Step 1: Calculate the concentration of acetic acid before ionization. 1 mol acetic acid g acetic acid " = 9.33 " 10 mol acetic acid g acetic acid 9.33 " 10 mol L soln = M acetic acid Step : We ignore water's contribution to [H ]. We consider CH 3 COOH as the only source of H ions. Step 3: Letting be the equilibrium concentration of H and CH 3 COO ions in mol L!1, we summarize: CH 3 COOH(aq) H (aq) CH 3 COO (aq) Initial (M): Equilibrium (M): Step 4: Write the ionization constant epression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (K a ), solve for. You can look up the K a value in Table 11.3 of the tet. K a! [H ][CH 3 COO " ] [CH 3 COOH] ( )( ) 1.8 " 10 = (0.0187! ) At this point, we can make an assumption that is very small compared to Hence, ( )( ) 1.8 " 10 = = M = [H ] = [CH 3 COO ] [CH 3 COOH] = ( )M = M Check: Testing the validity of the assumption, 5.8 " The assumption is valid. " 100% = 3.1% < 5% (a) Set up a table showing initial and equilibrium concentrations. C 6 H 5 COOH(aq) H (aq) C 6 H 5 COO (aq)

3 Initial (M): Equilibrium (M): (0.0 ) Using the value of K a from Table 11.3 of the tet: K a! [H ][C 6 H 5 COO " ] [C 6 H 5 COOH] 6.5 " 10 = (0.0! ) We assume that is small so (0.0 ) " 10 = 0.0 = M = [H ] = [C 6 H 5 COO ]! " 10 M Percent ionization = " 100% = 1.8% 0.0 M Set up a table as above. C 6 H 5 COOH(aq) H (aq) C 6 H 5 COO (aq) Initial (M): Equilibrium (M): ( ) Using the value of K a from Table 11.3 of the tet: K a! [H ][C 6 H 5 COO " ] [C 6 H 5 COOH] 6.5 " 10 = (0.0000! ) In this case we cannot apply the approimation that ( ) (see the discussion in Eample 11.11of the tet). We obtain the quadratic equation: ( ) ( ) = 0 The positive root of the equation is = M. (Is this less than 5% of the original concentration, M? That is, is the acid more than 5% ionized?) The percent ionization is then: 8.6 " 10 M Percent ionization = " 100% = 43% M All the listed pairs are ooacids that contain different central atoms whose elements are in the same group of the periodic table and have the same oidation number. In this situation the acid with the most electronegative central atom will be the strongest.

4 (a) H SO 4 > H SeO 4. H 3 PO 4 > H 3 AsO The salt ammonium chloride completely ionizes upon dissolution, producing 0.4 M [NH 4 ] and 0.4 M [Cl ] ions. NH 4 will undergo hydrolysis because it is a weak acid (NH 4 is the conjugate acid of the weak base, NH 3 ). Step 1: Epress the equilibrium concentrations of all species in terms of initial concentrations and a single unknown, that represents the change in concentration. Let ( ) be the depletion in concentration ( mol L!1 ) of NH 4. From the stoichiometry of the reaction, it follows that the increase in concentration for both H 3 O and NH 3 must be. Complete a table that lists the initial concentrations, the change in concentrations, and the equilibrium concentrations. NH 4 (aq) H O(l) NH 3 (aq) H 3 O (aq) Initial (M): Equilibrium (M): (0.4 ) Step : You can calculate the K a value for NH 4 from the Kb value of NH 3. The relationship is or K a K b = K w K! 14 Kw 1.0 " 10 a K b 1.8 " 10 = = = 5.6 " 10! 10 Step 3: Write the ionization constant epression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant (K a ), solve for. K a! [NH 3 ][H 3 O ] [NH 4 ]! " 10 = # 0.4! 0.4 = [H ] = M ph = log( ) = 4.8 Since NH 4 Cl is the salt of a weak base (aqueous ammonia) and a strong acid (HCl), we epect the solution to be slightly acidic, which is confirmed by the calculation The most basic oides occur with metal ions having the lowest positive charges (or lowest oidation numbers). (a) Al O 3 < BaO < K O CrO 3 < Cr O 3 < CrO We can write two equilibria that add up to the equilibrium in the problem. CH 3 COOH(aq) H (aq) CH 3 COO (aq) K = K a = 1.8! 10 "5

5 H (aq) NO (aq) HNO (aq) K = K a!1 = CH 3 COOH(aq) NO (aq) CH 3COO (aq) HNO (aq) 1 =. " 103!4 4.5 " 10 The equilibrium constant for this sum is the product of the equilibrium constants of the component reactions. K = ( )( ) = SO (g) H O(l) H (aq) HSO 3 (aq) Recall that 0.1 ppm SO would mean 0.1 parts SO per 1 million (10 6 ) parts of air by volume. The number of particles of SO per volume will be directly related to the pressure. 0.1 parts SO 7 SO atm 1. 10! P = = " atm 6 10 parts air We can now calculate the [H ] from the equilibrium constant epression [see Table 11.6 and associated footnote in the tet]. K! [H ][HSO 3 " ] P SO! 1.3 " 10 = 1. " 10! 7 = ( )( ) = M = [H ] ph = log( ) = 4.40