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050 moles) – (0. 030 moles) = zero. 020 moles of HS–. Now we will use the Henderson-Hasselbalch expression to discover the pH. we will use the variety of moles we simply calculated (0. 030 for S2– and nil. 020 for HS–) rather than concentrations simply because in an answer the concentrations are proportional to the variety of moles. pH = pK + log pK = –log(1. three × 10–13) = 12. nine pH = 12. nine + zero. 18 = thirteen. 1. (d) The response (H2S 2 H+ + S2–) is the sum of the 2 acid dissociation reactions less than. H2S H+ + HS– HS– H+ + S2– If one response is the sum of 2 different reactions, then its equilibrium consistent stands out as the made from the equilibrium constants for the opposite reactions. So Keq = K1K2 = (1. zero × 10–7)(1. three × 10–13) = 1. three × 10–20 3. (a) pKb = –log Kb = –log(1. seventy nine × 10–5) = four. seventy five (b) it is a buffered answer, so we’ll use the Henderson-Hasselbalch expression. First let’s locate [NH4+] and [NH3]. Moles = (molarity)(volume) Moles of NH4+ = (0. a hundred M)(0. a hundred L) = zero. 010 moles Moles of NH3 = (0. 200)(0. 080 L) = zero. 016 moles after we combine the strategies, the quantity turns into (0. a hundred L) + (0. 080 L) = zero. a hundred and eighty L. Molarity = [NH4+] = = zero. 056 M [NH3] = = zero. 089 M Now we will use the Henderson-Hasselbalch expression for bases. pOH = pK + log pK = four. seventy five pOH = four. seventy five + (–0. 20) = four. fifty five pH = 14 – pOH = 14 – four. fifty five = nine. forty five (c) First let’s learn how many moles of NaOH have been additional. Moles = Moles of NaOH = = zero. 005 mol while NaOH is extra to the answer, the subsequent response happens: NH4+ + OH– → NH3 + H2O So for each unit of NaOH further, one ion of NH4+ disappears and one molecule of NH3 seems. we will be able to use the result of the molar calculations we did partially (b). Moles of NH4+ = (0. 010) – (0. 005) = zero. 005 moles Moles of NH3 = (0. 016) + (0. 005) = zero. 021 moles Now we will use the Henderson-Hasselbalch expression. we will be able to use the variety of moles we simply calculated (0. 005 for NH4+ and zero. 021 for NH3) rather than concentrations simply because in an answer the concentrations could be proportional to the variety of moles. pOH = pK + log pK = four. seventy five pOH = four. seventy five + (–0. sixty two) = four. thirteen pH = 14 – pOH = 14 – four. thirteen = nine. 87 (d) while equivalent amounts of a base (NH3) and its conjugate acid (NH4+) are jumbled in an answer, the pOH may be equivalent to the pKb. From (a), pOH = pKb = four. seventy five pH = 14 – pOH = 14 – four. seventy five = nine. 25 ESSAYS 4. (a) An amphoteric species can act both as an acid or base. H2PO is amphoteric. Conjugate base: HPO Conjugate acid: H3PO4 or HPO is amphoteric. Conjugate base: PO Conjugate acid: H2PO (b) The smaller the dissociation consistent, the extra tightly the acid holds its hydrogen ions and the weaker the acid turns into. After a hydrogen ion has been got rid of, the remainder species has a detrimental cost that pulls the remainder hydrogen ions extra strongly. (c) H2PO4– H+ + HPO K2 = 6. 2 × 10–8 For H2PO4–, pKa = –log(6. 2 × 10–8) = 7. 2 the simplest buffer resolution is made with an acid whose pKa is nearly equivalent to the specified pH.
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