Chemistry: chapter 14 (first set of 15)

What are amphioprotic species? illustrate with suitable equations

These are substances that can donate a proton H+ and thus can act as an acid or a base: water acting as an acid: H2O(l) + NH3 (aq) (reversed arrows) NH4+(aq) + OH- (aq) water acting as a base: H20(l) + HCl (aq) (reversed arrows) H3O+ (aq) + Cl- (aq)

Show by net ionic equations that each of the following species can act as a Bronsted-Lowry Base: a. H20 b. OH- c. NH3 d. CN- e. S2- f. H2PO4-

H20: H20 (l) + H+ (aq) (reversed arrows) H3O+ (aq) OH-: OH- (aq) + H+ (aq) (reversed arrows) H2O (l) NH3: NH3 (aq) + H+ (aq) (reversed arrows) NH4+ (aq) CN-: CN- (aq) + H+ (aq) (reversed arrows) HCN (aq) S2-: S2- (aq) + H+ (aq) (reversed arrows) HS- (aq) H2PO4-: H2PO4- (aq) + H+ (reversed arrows) H3PO4 (aq)

Show by suitable net ionic equations that each of the following species an act as a Bronsted- Lowry acid: a. H3O+ b. HCL c. NH3 d. CH3COOH e. NH4+ F. HSO4-

H3O+: H30+ (aq) (reversed arrows) H+ (aq) + H2O (l) HCl: HCl (l) (reversed arrows) H+ (aq) + Cl- (aq) NH3: NH3 (aq) (reversed arrows) H+ (aq) + NH2- CH3COOH: CH3COOH (aq) (reversed arrows) H+ (aq) + CH3COO- NH4+: NH4+ (aq) (reversed arrows) H+ (aq) + NH3 (aq) HSO4-: HSO4- (aq) (reversed arrows) H+ (aq) + SO42- (aq)

Show by suitable net ionic equations that each of the following species can act as a Bronsted-Lowry acid: a. HNO3 b. PH4+ c. H2S d. CH3CH2COOH e. H2PO4- f. HS-

HNO3: HNO3 (aq) (reversed arrows) H+ (aq) + NO3- (aq) PH4+: PH4+ (aq) (reversed arrows) H+ (aq) + PH3 (aq) H2S: H2S (g) (reversed arrows) H+ (aq) + HS- (aq) CH3CH2COOH: CH3CH2COOH (aq) (reversed arrows) H+ (aq) + CH3CH2COO- (aq) H2PO4-: H2PO4- (aq) (reversed arrows) H+ (aq) + HPO42- (aq) HS-: HS- (aq) (reversed arrows) H+ (aq) + S2- (aq)

Show by suitable net ionic equations that each of the following species can act as a Bronsted- Lowry base: a. HS- b. PO43- c. NH2- d. C2H5OH e. O2- f. H2PO4-

HS-: HS- (aq) + H+ (aq) (reversed arrows) H2S (g) PO43-: PO43- (aq) + H+ (aq) (reversed arrows) HPO42- (aq) NH2-: NH2- (aq) + H+ (aq) (reversed arrows) NH3 (aq) C2H5OH: C2H5OH (aq) + H+ (aq) (reversed arrows) C2H5OH2 + (aq) O2-: O2- (aq) + H+ (aq) (reversed arrows) HO- (aq) H2PO4-: H2PO4- (aq) + H+ (aq) (reversed arrows) H3PO4 (aq)

Is the self-ionization of water endothermic or exothermic? The ionization constant for water (kw) is 2.9x10-14 at 40 degrees C and 9.6x10-14 at 6 degrees C.

Here, the value of kw increases with a rise in temperature. This means that as heat is added, concentration of the product increases. So, the self ionization of water is an endothermic process.

State which of the following species are amphioprotic and write chemical equations illustrating the amphioprotic character of these species: a. H2O b. H2PO4- c. S2- d. CO32- e. HSO4-

a. H2O: Here, H2O has the ability to donate a proton as well as accept a proton so it shows amphioprotic character: H2O(l) (YIELD) H+ (aq) + OH- (aq) H2O(l) + H+ (aq) (YIELD) H3O+ (aq) b. H2PO4-: Here, H2PO4- has the ability to donate a proton as well as accept a proton so it shows amphioprotic character: H2PO4- (aq) (YIELD) H+ (aq) + HPO42- (aq) H2PO4- (aq) + H+ (aq) (YIELD) H3PO4 (aq) c. S2-: Here, S2- has no H attached to it so it cannot donate a proton. so, it is not an amphioprotic species. d. CO32-: Here, CO32- has no H attached to it so it cannot donate a proton. So, it is not an amphioprotic species. e. HSO4-: Here, HSO4- has the ability to donate a proton as well as accept a proton. So, it shows amphioprotic character: HSO4- (aq) (YIELD) H+ (aq) + SO42- (aq) HSO4- (aq) + H+ (aq) (YIELD) H2SO4 (aq)

What is the conjugate acid of each of the following? What is the conjugate base of each? a. H2S b. H2PO4- c. PH3 d. HS- e. HSO3- f. H3O2+ g. H4N2 h. CH3OH

a. H2S: conjugate acid: H3S+, formed by accepting a proton conjugate base: HS-, formed by releasing a proton b. H2PO4-: conjugate acid: H3PO4, formed by accepting a proton conjugate base: HPO42-, formed by releasing a proton c. PH3: conjugate acid: PH4+, formed by accepting a proton conjugate base: PH2-, formed by releasing a proton d. HS-: conjugate acid: H2S, formed by accepting a proton conjugate base: S2-, formed by releasing a proton e. HSO3-: conjugate acid: H2SO3, formed by accepting a proton conjugate base: SO32-, formed by releasing a proton f. H3O2+: conjugate acid: H4O22+, formed by releasing a proton conjugate base: H2O2, formed by releasing a proton g. H4N2: conjugate acid: H5N2+, formed by accepting a proton conjugate base: H3N2-, formed by releasing a proton h. CH3OH: conjugate acid: CH3OH2+, formed by accepting a proton conjugate base: CH3O-, formed by releasing a proton

State which of the following species are amphioprotic and write balanced chemical equations illustrating the amphioprotic character of the species: a. NH3 b. HPO42- c. Br- d. N2 e. ASO43-

a. NH3: NH3 (aq) + H3O+ (aq) (YIELD) NH4OH (aq) + H2O (l) NH3 (aq) + OCH-3 (aq) (YIELD) NH2- (aq) + CH3OH b. HPO42-: Here, HPO42- has the ability to donate a proton as well as accept a proton. So, it shows amphioprotic character: HPO42- (aq) + OH (aq) (YIELD) H2O (l) + PO43- (aq) HPO42- (aq) + HClO4- (YIELD) H2PO4- (aq) + ClO4- c. Br-: Here, Br- has no attached H to it so it cannot donate a proton. So it is not an amphioprotic species. d. N2: Here, nitrogen is present in its highest oxidation state in NH4. It cannot accept one more proton so it is not an amphioprotic species. e. ASO43-: Here, ASO43- has no attached H to it. So, it cannot donate a proton so it is not an amphioprotic species.

Identify and label the Bronsted- Lowry acid, it's conjugate base, the Bronsted-Lowry base, and it's conjugate acid in each of the following equations: a. NO2- + H2O (YIELD) HNO2 + OH- b. HBr + H2O (YIELD) H3O+ + Br- c. HS- + H2O (YIELD) H2S + OH- d. H2PO4- + OH- (YIELD) HPO42- + H2O e. H2PO4- + HCl (YIELD) H3PO4 + Cl- f. [Fe(H2O)5(OH)]2+ + [Al(H2O)6]3+ (YIELD) [Fe(H2O)6]3+ + [Al(h2o)5(OH)]2+ g. CH3OH + H- (YIELD) CH3O- + H2

a. NO2- + H2O (YIELD) OH- + HNO2-switches OH- AND HNO2- BB BA CB CA b. HBr + H2O (YIELD) H3O+ + Br- BA BB CA CB c. HS- + H2O (YIELD) OH- + H2S-switches OH- and H2S BB BA CB CA d. H2PO4- + OH- (YIELD) H2O + HPO42- switches H2O and HPO42- BA (CB), BB(CA), CA, CB e. H2PO4- + HCL (YIELD) H3PO4 + Cl-switches Cl- and H3PO4 BB (CA), BA(CB), CB, CA f. [Fe(H20)5(OH)2+ + [Al(H2O)6]3+ (YIELD) [Al(H2O)5(OH)]2+ + [Fe(H2O)6]3+-switches al and fe combos BB (CA), BA(CB), CB, CA g. CH3OH + H- (YIELD) H2 + CH3O- BA BB CA CB switiches H2 and CH3O-

What is the conjugate acid of each of the following? What is the conjugate base of each? a. OH- b. H20 c. HCO3- d. NH3 e. HSO4- f. H2O2 g. HS- h. H5N2

a. OH-: conjugate acid: H2O, formed by accepting a proton. conjugate base: 02-, formed by releasing a proton b. H2O: conjugate acid: H3O+, formed by accepting a proton. conjugate base: OH-, formed by releasing a proton. c. HCO3-: conjugate acid: H2CO3, formed by accepting a proton. conjugate base: CO32-, formed by releasing a proton. d. NH3: conjugate acid: NH4+ formed by accepting a proton conjugate base: NH2-, formed by releasing a proton e. HSO4-: conjugate acid: H2SO4-, formed by accepting a proton. conjugate base: SO42-, formed by releasing a proton f. H2O2: conjugate acid: H3O2+, formed by accepting a proton. conjugate base: HO2-, formed by releasing a proton g. HS-: conjugate acid: H2S, formed by accepting a proton conjugate base: S2-, formed by releasing a proton h. H5N2: conjugate acid: H6N22+, formed by accepting a proton conjugate base: H4N2, formed by releasing a proton

Identify and label the Bronsted- Lowry acid, it's conjugate base, the Bronsted-Lowry base and its conjugate acid in each of the following equations: a. HNO3 + H2O (YIELD) H3O+ + NO3- b. CN- + H2O (YIELD) HCN + OH- c. H2SO4 + Cl- (YIELD) HCl + HSO4- d. HSO4- + OH- (YIELD) H2O + SO42- e. O2- + H2O (YIELD) OH- + OH- f. [CU(H2O)3OH]+ + [Al(H2O)6]3+ (YIELD) [CU(H2O)4]2+ + [Al (H2O)5(OH)3+ g. NH2- + H2S (YIELD) HS- + NH3

a. [HNO3 (BA), NO3- (CB)] [H2O (BB), H3O+ (CA) b. CN- + H2O (YIELD) OH- + HCN-switches oh- and hcn- [BB (BA (CB [CA c. H2SO4 + Cl- (YIELD) HCl + HSO4- [BA (BB (CA [CB d. HSO4- + OH- (YIELD) H2O + SO42- [BA (BB (CA [CB e. O2- + H2O (YIELD) OH- + OH- [BB (BA (CB [CA f. BB (CB), BA(CB), CA, CB g. NH2- + H2S (YIELD) HS- + NH3 BB BA CB CA

Write equations to show NH3 as both a conjugate base and a conjugate acid

as a conjugate acid: NH2- + (reversed arrows) NH3 as a conjugate base: NH4+ (reversed arrows) H+ +NH3

Write equations that show H2PO4- acting both as an acid and as a base

as an acid: H2PO4- (reversed arrows) H+ + HPO42- as a base: H2PO4- + H+ (reversed arrows) H3PO4

Explain why a sample of pure water at 40 degrees C is neutral even though [H3O+] = 1.7x10-7M. Kw is 2.9x10-14 at 40 degrees C.

kw=[H30+][OH-] [OH-] = kw/[H3O+] = 2.9x10-14/1.7x10-7M = 1.7x10-7M as [H3O+]= [OH-] so, the given sample of pure water is neutral