K w : K w = 1 25 C K w = K a K b 14 = ph + poh. Henderson- Hasselbalch Equation; ph = pk a + log [C b /C a ] or poh = pk b + log [C a /C b ]

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1 Quiz 2 Name Key (last) (First-Name) Analytical Chemistry October 21, 2018 By signing your signature above you agree that you have worked alone and neither give nor received help from any source. You must upload this in Blackboard for full credit. Do not share your answers with anybody. You score on this exercise will only give you a maximum of 60% of what you missed in the second quiz. You must show your complete work to receive any points px and [X] Relationship ph = - [H 3 O + ] poh = - [OH - ] pk a = - [K a ] [H 3 O + ]= -ph [OH - ]= -poh [K a ]= -pka K w : K w = 25 C K w = K a K b 14 = ph + poh Henderson- Hasselbalch Equation; ph = pk a + [C b /C a ] or poh = pk b + [C a /C b ] pka for various substances: Table 1 will be useful for several problems in this exam Substance Formula pk a1 pk a2 pk a3 i) Phenol, C 6 H 5 OH 9.89 ii) Hydrogen cyanide, HCN 9.32 iii) Ammonium, NH iv) Hypobromous acid HBrO 8.68 v) Hypochlorous acid HClO 7.52 vi) Hydrogen sulfide H 2 S vii) rbonic acid H 2 CO viii) Nicotinic acid HC 6 H 4 NO ix) Acetic acid HC 2 H 3 O x) Barbituric acid HC 4 C 3 N 2 O xi) actic HC 3 H 5 O xii) Formic acid HCHO xiii) Nitrous acid HNO xiv) Hydrofluoric acid HF 3.17 xv) Chloroacetic acid HC 2 H 2 O 2 Cl 2.85 xvi) Phosphoric acid H 3 PO xvii) Arsenic acid H 3 AsO In any aqueous solution Kw is temperaturedependent, the auto-ionization rxn is endothermic, so Kw increases with temperature ( C) xviii) Sulfurous acid H 2 SO xix) Sulfuric acid H 2 SO 4 Strong 1.99 T + 32 F =1.8T C

2 1 lculate the activity coefficient of X2+ when µ = using the following method: i) Use the extended Debye-Huckel equation (eqn 7-6). ii) Use Table 7-1 using linear interpolation. 2 lculate ppb when of 0.0 M CO32 is titrated with of 0.0 M Pb2+. Ksp = 7.4 x 14 for PbCO3. 3 Match the following charge solution containing chemical specie to the correct charge balance equation. i. Solution containing: H+, OH -, 2+, HCO3-, CO32-, (HCO3)+, (OH)+, K+, and ClO4[H+] + 2 [2+] + [(HCO3)+] + [(OH)+] + [K+] = [OH - ] + [HCO3- ] + 2 [CO32- ] + [ClO4- ] ii. Solution of H2SO4 in water if the H2SO4 ionizes to HSO4- and SO42[H+] = [OH - ] + [HSO4- ] + 2 [SO42- ] iii. Solution of arsenic acid, H3AsO4, in which the acid can dissociate to H2AsO4-, HAsO42-, and AsO43[H+] = [OH- ] + [H2AsO4- ] + 2 [ HAsO42- ] + 3 [AsO43- ]

3 4 lculate the ph of the solution describe with and without activity. What is the ph without and with activity for a 0.250M NaF solution in 0.05M NaCl? a) Without activity? 5 b) With activity? Consider the titration of 25.0 of M MnSO4 with 0.00 M EDTA in a solution buffered to ph lculate pmn2+ at the following volumes of added EDTA and sketch the titration curve. Use two significant figures when typing your pmn2+ answer. VO pmn2+ Reaction: Mn + EDTA D MnY K' = a K = (4.2 ) ( ) = 3.3 The eq point is i) 0.00 ml: f Y f 0.0: [Mn+2] = M, pmn2+ = -[0.0200] pmn2+ = 1.70 ii) ml: : [Mn+2] = [0.50m Mn m EDTA] / 44 = M pmn2+ = 2.15 iii) ml: : [Mn2+ ] = [0.50m Mn m EDTA] / 54 = M pmn2+ = 2.41 iv) ml: : [Mn2+ ] = [0.50m Mn m EDTA] / 74.9 = M pmn2+ = 4.87 v) ml: 6.87 vi) 50. ml: 8.86 vii) 52.0 ml:.16 viii) 65.0 ml:.04: 50.00: [0.500 m Mn2+ ] [0.500 m MnEDTA2- ] (0.500 m MnEDTA2- ] / 75 = M Mn2+ + EDTA! MnY-2 x x x ( x) / x2 = a 4- K f = 3.3 x = pmn2+ = 6.85 Y m EDTA or 0.01 m excess EDTA after 5.00 m Mn+2 reaction [EDTA] = [0. / 75.1] (0.00 M) = M = [25.0 / 75.1] ( M) = M = M pmn2+ = m EDTA or m excess EDTA after 5.00 m Mn+2 reaction [EDTA] = [2 / 77 ] (0.00 M) = M = [25.0 / 77] ( M) = M = M pmn2+ = m EDTA or 0.15 m excess EDTA after 5.00 m Mn+2 reaction [EDTA] = [15 / 90 ] (0.00 M) = M = [25.0 / 90] ( M) = M = M pmn2+ =.00

4 6 Titration of weak acid with strong base ink to pdf hardcopy. Download the pdf and answer the question, take a photo of your work and upload it with the link below (instead of the dropbox). If you do not show your work, you will not receive credit for this question. If you have multiple pages, be sure to zip it to one file. You can also copy and past the photos into word and send me one file. lculate the ph values for a titration curve when 15.0 of M Formic acid is titrated with a calcium hydroxide solution that was prepared by taking 18.5 g (OH)2 and dissolved in 500- solution. lculate the ph at 0.00%, 50.0%, 90.0% 0.% and 0.% of the equivalent point. (OH) 2 = g/, 18.5 g * (1/ g) * (1/.500) = 0.500M [OH-] = 1.0M ( M) M = 7.50 m titrated with 1.0M OH- 0% = 0, 50% = 3.75, 90% = 6.75, 0% eq pt = 7.50, 0% = 8.25 Vtotal = % and Vol Equation to use Hypobromos acid Nicotinic actic Formic Nitrous pk a = K a = Pk b = K b = %, 0 x 2 ph = * k a = Solve via 0.500M x = [H 3 O+ ] = k a 0.500M Quadratic 50%, 3.75! ph = ph=pk a + C # b & # C &, ph = pk a a % 90%, 6.75 HA + OH - H 2 O + A - ph = i m 9m 0%, %, 8.25 e 1 m 9m! ph=pka+ C # b & 1m, HA = A - = 9m # C & a % A - + H 2 O HA + OH - i m 30ml Δ - x + x + x e 1m / 31 = [OH-] = M, poh = 1.492, ph = Except for Hypobromos acid, [OH-] = M x = [OH - ] = k b 0.333M ph =.0* Solve via Quadratic ph = OH- = OH- 1 + OH

5 7 You want to prepare a buffered solution with a ph = (Below are two methods to prepare this solution) i) What mass in gram of NaF must be added to solution of M HF? ph = pka +, Molar Mass (NaF) = = = g!3.00# = 2.138, = (3.00) = M Mass NaF = = =0.33! -# F & '! # &HF' g Mass NaNO 2 =269.38g = 269g = 0.33 = ph = pka = , Molar Mass (NaF) = = g = =0.33! -# F & ' & ' = 0.33 = = !NO- # 2 = ,!NO-2# = (3.50) = M!3.50# 69.1g Mass NaNO 2 = Mass NaNO 2 =541.43g = 541g = !NO- # 2!HNO # 2 = = = 0.35,!x# & ' & ' = 0.35 = = 2.239!x# = x = x x = M Mass NaOH = g 1 = g NaOH