REVIEW QUESTIONS Chapter 16

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1 Chemistry 102 ANSWER EY REVIEW QUESTIONS Chpter A buffer is prepred by dding 20.0 g of cetic cid (HC 2 H O 2 ) nd 20.0 g of sodium cette (NC 2 H O 2 ) in enough wter to prepre 2.00 L of solution. Clculte the ph of this buffer? ( 1.8x10 5 ) 20.0 g 1 mol 1 1 mol 1 HAc x x M 20.0 g NAc x x M 60.0 g 2.00 L 82.0 g 2.00 L HC 2 H O 2 H 2 O H O C 2 H O 2 Initil x x x Equil x x x [HO ][C2HO2 ] (x)(0.122 x ) 1.8x10 5 [HC H O ] x (0.167)(1.8x10 5 ) x 2.6x ph log(2.6x10 5 ) Wht is the rtio of HCO to H 2 CO in blood of ph 7.? ( for H 2 CO.x10 7 ) H 2 CO H 2 O H O HCO [HCO ] [HCO ] [H CO ] [H2CO ] ph p log log 2 [HCO ] ntilog (7. [H CO ] ) How mny grms of NBrO should be dded to 1.00 L of M HBrO to form buffer with ph of 8.80? ( for HBrO 2.5x10 9 ) HBrO H 2 O H O BrO ph p log [BrO ] [BrO ] log [HBrO] [HBrO] [BrO ] ntilog ( ) [HBrO] [BrO ] 1.6 (0.200 M) 0.2 M 1.00 L x 0.2 mol x 1 L g 8 g 1 mol 1

2 . Acetylslicylic cid (irin, HC 9 H 7 O ) is wek cid with 2.75x10 5 t 25 C..00 g of sodium cetylslicylte (NC 9 H 7 O ) is dded to ml of M solution of this cid. Clculte the ph of the resulting solution t 25 C. Molrity of NC9H7O.00 g 1 mol 1 x x 0.07 M 202 g L HC 9 H 7 O H 2 O H O C 9 H 7 O Initil x x x Equil x x 0.07 x [HO ][C9H7O ] (x)(0.07 x ) 2.75x10 5 [HC H O ] x 9 7 (0.100)(2.75x10 5 ) x.70x ph log(.70x10 5 ) The equtions nd dissocition constnts for three different cids re given below: HCO H 2 PO HSO H CO 2 H HPO 2 H SO 2.2x10 7 p x10 8 p x10 2 p 1.9 Identify the conjugte pir tht is best for prepring buffer with ph of 7.2. Clerly explin your choice. The best conjugte pir would be H 2 PO nd HPO 2 The ph p 7.2 for this buffer when [H 2 PO ] [HPO 2 ] 2 [HPO ] ph p log [H PO ] 2 2

3 6. A smple of 25.0 ml of M solution of HBr is titrted with M NOH. Clculte the ph of solution fter 10.0 ml of the bse is dded. HBr NOH NBr H 2 O Initil 2.50 mmol 2.00 mmol mmol 2.00 mmol 2.00 mmol ---- Finl 0.50 mmol mmol mmol [H ][HBr] 0.01 M ph -log (0.01) ml 7. A buffer solution is prepred by dding 0.10 L of 2.0 M cetic cid solution to 0.10 L of 1.0 M NOH solution. ) Clculte the ph of this buffer solution L 0.10 L x x 2.0 mol 0.20 mol HC2HO 2 1 L 1.0 mol 0.10 mol NOH 1 L HC 2 H O 2 NOH NC 2 H O 2 H 2 O Initil Finl mol 0.10 mol ] 0.50 M [HC H O ] 0.50 M 0.20 L 0.2 L [C2HO2 ] ph p log [HC H O ] [C2HO2 From textbook 1.7x10 5 p ph.77 log

4 b) 0.10 L of 0.20 M HCl is dded to 0.0 L of the buffer solution bove. Wht is the ph of the resulting solution? The H O ions provided by HCl rect with the cette ions in the buffer. [HO ] (0.10L)(0.20 M) mol [C2HO2 ] [HC H O ] (0.0 L)(0.50 M) 0.20 mol C 2 H O 2 H O HC 2 H O 2 H 2 O Initil Finl mol 0.22 mol ] 0.6 M [HC H O ] 0. M 0.50 L 0.50 L [C2HO2 ] 5 ph p log From textbook 1.7x10 [HC H O ] [C2HO2 0.6 p.77 ph.77 log A 10.0 ml solution of M NH ( b 1.8 x10 5 ) is titrted with M HCl solution. Clculte the ph of this solution t equivlence point mol 1 HCl 1 L At equivlence point 10.0 ml NH x x x 10.0 ml of HCl 1 L 1 NH mol At equivlence point ll NH (1.00 mmol) rects with ll HCl (1.00 mmol) to produce 1.00 mmol of NH Cl. Since only slt is present t this point, the ph of solution is bsed on hydrolysis of this slt mmol [NH ] M 20.0 ml NH H 2 O NH H O Initil x ---- x x Equil x ---- x x 1 1.0x10 1.8x10 5.6x10 10 [HO ][NH ] (x)(x) x10 [NH ] x w b [HO ] x (0.050)(5.6x10 10 ) 5.x10 6 ph log(5.x10 6 ) 5.28

5 9. A 10.0-mL solution of 0.00 M NH is titrted with M HCl solution. Clculte the ph fter the following dditions of the HCl solution: () 0.0 ml, (b) 10.0 ml, (c) 0.0 ml ) Since no cid hs been dded, the ph of solution is bsed on the ioniztion of NH. NH H 2 O NH OH From textbook, b 1.8 x 10 5 b [NH ][OH ] (x)(x) [NH ] x 1.8x10 5 x [OH ] (0.00)(1.8x10 5 ) 2.2x10 poh -log(2.2x10 ) 2.6 ph b) Addition of 10.0 ml of cid neutrlizes some of the mmoni, s shown below: NH HCl NH Cl Initil.00 mmol 1.00 mmol mmol 1.00 mmol 1.00 mmol ---- Finl 2.00 mmol mmol mmol 1.00 mmol [NH ] M [NH ] M 20.0 ml 20.0 ml x x 10 p 5 log x 10 [bse] ph p log 9.25 log 9.55 [cid] c) After ddition of 0.0 ml of HCl equivlence point is reched. At this point ll NH (.00 mmol) rects with ll HCl (.00 mmol) to produce.00 mmol of NH Cl. Since only slt is present t this point, the ph of solution is bsed on hydrolysis of this slt. [NH ].00 mmol M 0.0 ml NH H 2 O NH H O Initil x ---- x x Equil x ---- x x [NH ][HO ] (x)(x) 5.56 x 10 [NH ] ( x ) x [HO ] (0.0750)(5.56x10 10 ) 6.6x10 6 ph log(6.6x10 6 )

6 10. A 5.0-mL smple of M cetic cid is titrted with M NOH. Clculte the ph of the solution () before ddition of NOH, (b) fter ddition of 20.0 ml of NOH nd (c) t the equivlence point. ) Since no bse hs been dded, the ph of solution is bsed on the ioniztion of cid. HC 2 H O 2 H 2 O C 2 H O 2 H O From textbook, 1.7 x 10 5 [CHO ][H O ] (x)(x) 2 [HC2HO 2] x 1.7x10 5 x [HO ] (0.200)(1.7x10 5 ) 1.8x10 ph log(1.8x10 ) 2.7 b) Addition of 20.0 ml of NOH neutrlizes some of the cetic cid, s shown below: HC 2 H O 2 NOH NC 2 H O 2 H 2 O Initil 9.00 mmol.60 mmol mmol.60 mmol.60 mmol ---- Finl 5.0 mmol 0.60 mmol mmol [HC2H O 2] M [C2H O ml 1.7 x mmol ] M 65.0 ml [bse] p log.77 ph p log.77 log.59 [cid] c) At equivlence point: mol 1 mol bse 1 L 5.0 ml cid x x x 50.0 ml of bse 1 L 1 mol cid mol At this point ll the cid (9.00 mmol) is neutrlized by the bse (9.00 mmol) to produce 9.00 mmol of slt. Since only slt is present, the ph of the solution is bsed on hydrolysis of this slt. [C2HO mmol ] M 95.0 ml C 2 H O 2 H 2 O HC 2 H O 2 OH Initil x ---- x x Equil x ---- x x x 10 b 1.7 x x 10 [C 10 2HO2 ][OH ] (x)(x) 5 b 5.88 x 10 [HC H O ] (0.097 x ) x [OH ] (0.097)(5.88x10 10 ) 7.7x10 6 poh log(7.7x10 6 ) 5.1 ph

7 11. Clculte the molr solubility of AgBr ( 5.0x10 1 ) in 0.50 M NBr solution. AgBr (s) Ag (q) Br (q) Initil x x x Finl x 0.50 x - [Ag ][Br ] (x)(0.50 x 5.0x ) 5.0x solubility x 1.0x10 M 12. A solution is mde by mixing 10.0 ml of 0.10 M Pb(NO ) 2 nd 10.0 ml of M N 2 SO. Will precipitte form? ( for PbSO 1.06x10 8 ) PbSO (s) Pb 2 (q) SO 2 (q) 2 (0.10 M)(10.0 ml) 2- ( M)(10.0 ml) - [Pb ] M [SO ] 5.00x10 M (20.0 ml) (20.0 ml) Q [Pb ][SO ] (0.0500)(5.00x10 ) 2.50x Since Q >, precipittion will occur The solubility of iron (II) hydroxide, Fe(OH) 2, is 1.x10 g/l. ) Clculte the for iron (II) hydroxide. Fe(OH) 2 (s) Fe 2 (q) 2 OH (q) 2 [Fe ] [Fe(OH) 2] - 1.x10 g 1 mol x 1 L g [OH ] 2 (1.59x10 ).18x10 M x10 M [Fe ][OH ] (1.59x10 )(.18x10 ) 1.61x b) Clculte ph of sturted solution of iron (II) hydroxide From prt () [OH ].18x10 M -1 w 1.0x10-10 [H ].1x10 M - -5 [OH ].18x10 ph -log [H ]

8 c) A 50.0 ml smple of.00x10 M FeSO solution is dded to 50.0 ml of.00x10 6 M NOH solution. Does precipitte form? - 2 (.00x10 M)(50.0 ml) - [Fe ] 1.50x10 M (100.0 ml) -6 - (.00x10 M)(50.0 ml) -6 [OH ] 2.00x10 M (100.0 ml) Q [Fe ][OH ] (1.50x10 )(2.00x10 ) 6.00x Since Q <, precipittion will not occur 1. Led iodte, Pb(IO ) 2, is slightly soluble slt with of 2.6x10 1 t 25 C. To 5.0 ml of Pb(NO ) 2 solution is dded 15.0 ml of M IO. A precipitte of Pb(IO ) 2 results. Wht re the [Pb 2 ] nd [IO ] in the finl solution? 2 (0.150 M)(5.0 ml) - (0.800 M)(15.0 ml) [Pb ] M [IO ] 0.20 M (50.0 ml) (50.0 ml) Using bounce-bck method, first ssume ll Pb 2 rects with ll IO ion to produce Pb(IO ) 2, nd then some of the precipitte dissolves bck to the ions. Pb(IO ) 2 (s) Pb 2 (q) 2 IO (q) Initil Precipitte x x 2x Finish x 0.002x [Pb ][IO ] (x)(0.00 x ) 2 2.6x x10-10 [Pb ] x 2.9x10 M 2 (0.00) [IO ] x 0.00 M - 8

9 15. Consider solution tht is M in B 2 nd M in C 2. If sodium sulfte is dded to this solution to selectively precipitte one of the ctions, which will precipitte first? Wht is the minimum concentrtion of N 2 SO tht would trigger the precipittion of this ction? From textbook, for BSO 1.07x10 10 for CSO 7.10x10 5 Since the solution stoichiometry for both these compounds re similr, it would be pproprite to relte nd molr solubility. Since the lower vlue would require the lower sulfte ion concentrtion in order to precipitte, it would then follow tht B 2 would precipitte first. BSO (s) B 2 (q) SO 2 (q) [B ][SO ] 1.07x x10-8 [SO ] 1.07x10 M 2 [B ] M 16. Wht is the Cr concentrtion when mol of Cr(NO ) is dissolved in liter of solution buffered t ph of Cr forms complex ion with hydroxide shown below: Cr (q) OH () Cr(OH) f 8 x10 29 poh [OH ] ntilog poh 1.0 x10 Due to the lrge f ll of Cr is converted to the complex ion, nd some subsequently dissocites bck to Cr. Then t equilibrium, Cr (q) OH () Cr(OH) x 1.0 x x - [Cr(OH) ] f - [Cr ][OH ] -16 [Cr ] x 1.2x10 M ( x ) - 8x10 x (1.0x10 ) 29 Note: Since the solution is buffered, [OH ] will remin constnt during the rection. 9

10 17. A 0.10-mol smple of AgNO is dissolved in 1.00 L of 1.00 M NH. If mol of NCl is dded to this solution, will AgCl ( 1.8 x10 10 ) precipitte? (Ag nd NH form the complex ion [Ag(NH ) 2 ] with f 1.6 x 10 7 ) To determine if precipitte forms, we need to determine the concentrtion of Ag in the solution, nd then clculte Q to determine if precipitte forms. To determine the concentrtion of Ag, due to the lrge f vlue, ssume ll of the Ag rects to form the complex nd then some dissocites bck. Ag (q) 2 NH (q) Ag(NH ) 2 (q) Initil Complex x 2x x Finish x 0.802x 0.10x [Ag(NH ) ] 2 f 2 [Ag ][NH ] ( x ) x (0.80 2x ) 2 1.6x [Ag ] x 9.8x10 M 2 7 (0.80) (1.6x10 ) 7 Q [Ag ][Cl ] (9.8x10 9 )(0.010) 9.8x10 11 < 1.8x10 10 AgCl will NOT precipitte 10

11 18. AgNO is dded to solution tht is 0.10 M in NCl nd M 2 CrO. Assume no dilution cused by the ddition of AgNO. Given the vlues below: for AgCl 1.6x10 10 for Ag 2 CrO 9.0x10 12 ) Which precipittes first, AgCl or Ag 2 CrO? Clculte the [Ag ] when precipittion first begins. Since the solution stoichiometry for both these compounds re not the sme, it would not be pproprite to relte nd molr solubility. To determine which ion precipittes first, we must clculte the ction concentrtion required for ech precipittion. The lower vlue required for precipittion would indicte the ion tht would precipitte first. [Ag ] required for AgCl precipittion: AgCl (s) Ag (q) Cl (q) [Ag ][Cl ] 1.6x x10-9 [Ag ] 1.6x10 M - [Cl ] 0.10 M [Ag ] required for Ag 2 CrO precipittion: Ag 2 Cr (s) 2 Ag (q) CrO 2 (q) [Ag ] [CrO ] 9.0x x10 [Ag ].0x10 M [CrO ] M The precipittion occurs for the slt which requires the smllest [Ag ] t equilibrium. Therefore AgCl precipittes first t [Ag ] of 1.6x10 9 M b) Wht is the [Cl ] when Ag 2 CrO first begins to precipitte? Clcultions bove show tht in order of Ag 2 CrO to precipitte, the [Ag ] must equl.0x10 5 M. Therefore, [Ag ][Cl ] (.0x10 ) [Cl ] 1.6x10 1.6x10 [Cl ].0x x10 M 11

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