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South Pasadena • AP Chemistry Name 4 ▪ Acid Base Solutions Period 4.1 PROBLEMS – TYPES OF ACIDS Date AND BASES 1. For each of the following, identify whether they are an acid, base, amphoteric, or neither. Write its conjugate and dissociation equation. Type Compound − Conjugate Dissociation Equation HBr (Br− is conjugate of SA) (a) Neither Br (b) Base C2H3O2− HC2H3O2 HC2H3O2 + H2O H3O+ + C2H3O2− (c) Acid C5H5NH+ C5H5N C5H5NH+ + H2O H3O+ + C5H5N (d) Base C6H5NH2 C6H5NH3+ C6H5NH2 + H2O C6H5NH3+ + OH− (e) Neither CH3OH (f) Base CN− HCN CN− + H2O HCN + OH− (g) Acid H2CO3 HCO3− H2CO3 + H2O H3O+ + HCO3− (h) Base H2NNH2 H2NNH3+ H2NNH2 + H2O H2NNH3+ + OH− (i) Amphiprotic HCO3− H2CO3 / CO32− HCO3− + H2O H2CO3 + OH− HCO3− + H2O CO32− + H3O+ (j) Acid HF F− HF + H2O H3O+ + F− (k) Acid HNO3 NO3− HNO3 → H+ + NO3− (l) Acid HOCl OCl− HOCl + H2O H3O+ + OCl− (m) Neither K+ (n) Acid NH4+ NH3 NH4+ + H2O H3O+ + NH3 (o) Base NO2− HNO2 NO2− + H2O HNO2 + OH− (p) Base Zn(OH)2 (Alcohols are neither acids nor bases) (K+ is the cation of a SB) 2. The monohydrogenphosphate ion, HPO42−, is amphiprotic. Write the chemical equation for the ion behaving as an acid and as a base. Identify on the equation the acid, base, conjugate acid, and conjugate base. HPO42− + H2O H3O+ + PO43− Acid Base Conj Acid Conj Base Zn(OH)2 Zn2+ + 2 OH− 4. Write the net ionic equations for the acid-base reactions between: a. Hydrofluoric acid and sodium hydroxide HF + OH− → F− + H2O b. Ammonium chloride and potassium hydroxide NH4+ + OH− → NH3 + H2O HPO42− + H2O H2PO4− + OH− Base Acid Conj Acid Conj Base 3. Arsenic acid, H3AsO4, is a triprotic acid. Write the three dissociation equations for this acid. H3AsO4 + H2O H3O+ + H2AsO4− H2AsO4− + H2O H3O+ + HAsO42− HAsO42− + H2O H3O+ + AsO43− c. Sodium bicarbonate and sulfuric acid HCO3− + H+ → H2O + CO2 d. Chlorous acid and aqueous ammonia solution HClO2 + NH3 ClO2− + NH4+ e. Disodium hydrogen phosphate and acetic acid HPO42− + HC2H3O2 H2PO4− + C2H3O2− 5. Write the balanced molecular equations for the neutralization reactions to form each salt. a. Calcium bromate 2 HBrO3 + Ca(OH)2 → Ca(BrO3)2 + 2 H2O 8. Without using a calculator, determine the whole number pH between which the solution lies. a. A solution with pOH = 4.2 has a pH between 9.0 and 10.0. b. Aluminum bisulfate 3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O b. A solution with [H+] = 8.9 × 10−8 M has a pH between 7.0 and 8.0. c. Monosodium dihydrogen phosphate H3PO4 + NaOH → NaH2PO4 + H2O c. A solution with [OH−] = 4.1 × 10−1 M has a pH between 13.0 and 14.0. 6. A neutral amino acid is amphoteric. Draw the structures of its conjugate acid and its conjugate base. H N H H O C C O H [H+] 10. Consider the ratio: [OH−] . R Conjugate acid: H + N H H H O C C O H R Conjugate base: H N H H O C C O R 7. Many salts are acidic or basic when dissolved in water. To determine the acidity of a salt, examine whether each ion is acidic, basic, or neutral. For example, NaF is a basic salt because Na+ is neutral while F− is basic. For each of the following salts, determine whether its acidic, basic, or neutral in solution. a. NH4Cl Acidic (NH4+) b. KNO3 Neutral c. Na2CO3 Base (CO32−) + d. C2H5NH3Br Acid (C2H5NH3 ) e. LiCN Base (CN−) f. Base (C2H3O2−) CsC2H3O2 g. RbClO4 9. At 0°C, the pH of pure water is 7.47. Find the value of Kw at this temperature. [H+] = [OH−] = 10−7.47 = 3.4 × 10−8 M Kw = [H+][OH−] = (3.4 × 10−8)2 = 1.1 × 10−15 Neutral a. What is the value of the ratio for: [H+] A neutral solution? =1 [OH−] [H+] Acidic solution? >1 [OH−] [H+] Basic solution? <1 [OH−] b. Kw of pure water at 40°C is 2.92 × 10−14. Consider a 0.0010 M solution of HCl. What is the [OH−] at: 25°C Kw 1.0 × 10−14 [OH−] = + = [H ] 0.0010 = 1.0 × 10−11 M [H+] 1.0 × 10−3 = 1.0 × 108 − = [OH ] 1.0 × 10−11 40°C Kw 2.92 × 10−14 [OH−] = + = [H ] 0.0010 −11 = 2.9 × 10 M + [H ] 1.0 × 10−3 = = 3.4 × 107 [OH−] 2.9 × 10−11 c. Does the solution become more acidic, become more acidic, or remain unchanged as [H+] the temperature increases? Use the [OH−] ratio to justify your answer. The solution became less acidic as the [H+] temperature increased because the [OH−] ratio decreased.