COMMON ION & BUFFER PROBLEMS KEY

Transcription

1 COMMON ION & BUFFER PROBLEMS KEY 1) What is the ph of a solution containing 0.0 M NH and 0.15 M NH 4 NO? K b for NH = NH is a weak base: NH H O NH 4 OH - NH 4 NO is a salt: NH 4 NO NH 4 NO ; thus NH 4 is a common ion NH H O NH 4 OH - [NH M [H O [NH 4 M [OH - M I C - E K b = [ NH 4 [ NH [ OH Approimation: ignore, terms: = ( ) ( 0. 0) = [OH - = M poh = -log = 4.44 ph = = 9.56 ph = 9.56 (This problem can also be solved using the K a rn: NH 4 NH H ; if you use this reaction, you must convert K b to its corresponding K a value.) ) A buffer solution contains 0.0 M HCHO and 0.0 M NaCHO. The volume of the solution is 15 ml. K a for HCHO = a) What is the ph of this buffer solution? Salt: NaCHO Na - CHO Acid ionization rn: HCHO H - CHO I C - E Approimation: ignore, terms [ H [ CHO [ HCHO = ( 0.0) ( 0.0) = [H = ph = -log[h = -log =.9 ph =.9 b) If 50.0 ml of 0.10 M NaOH is added to the buffer solution, what is the ph? Strong base: NaOH Na OH - diluted so recalculate M: M HCHO = ( 0.0 M )( 15 ) (175) (175) = 0.14 M M CHO - = ( 0.0 M )( 15 ) = 0.1 M; M OH - = ( 0.10 M )( 50.0 ) (175) = 0.09 M Buffer, Titration and Solubility problems Key 1

2 neutralization reaction: OH - - HCHO CHO Initial Change Final Acid ionization rn: HCHO H - CHO I C - E [ H [ CHO [ HCHO = ( 0.4) ( 0.11) H O = [H = ph = -log[h = -log = 4.08 ph = 4.08 *For a buffer solution, ph only rises a little if a small amount of strong base is added. c) If 50.0 ml of 0.10 M HCl is added to the buffer solution, what is the ph? Strong acid: HCl H O H O Cl - diluted so recalculate M: M HCHO = ( 0.0 M )( 15 ) (175) neutralization reaction: H - CHO (175) = 0.14 M M CHO - = ( 0.0 M )( 15 ) = 0.1 M; M H = ( 0.10 M )( 50.0 ) HCHO Initial Change Final Acid ionization rn: HCHO H - CHO I C - E [ H [ CHO [ HCHO = ( 0.18) ( 0.17) (175) = 0.09 M = [H = ph = -log[h = -log =.77 ph =.77 * For a buffer, ph only drops a little when a small amount of strong acid is added. Buffer, Titration and Solubility problems Key

3 TITRATION PROBLEMS KEY 1. A 0.00 sample of M HCl is titrated with 0.00 M NaOH. Calculate the ph of the solution after the following volumes of NaOH have been added: a) 0 ml; b) ml; c) 15.0 ml; d) 0.00 ml. a) 0 of NaOH added only SA is present initially: For strong acid: [H = [HCl = M HCl ph = -log[h = -log(0.150) = 0.84 b) of NaOH neutralization reaction: HCl NaOH NaCl H O SA SB moles HCl moles HCl = 0.00 = moles HCl m 0.00 moles NaOH moles NaOH = = moles NaOH m After neutralization: moles ecess acid = moles moles = moles HCl M H moles = M HCl = = 0.0 M L ph = - log [H = - log 0.0 = c) 15.0 ml of NaOH From part b, moles HCl = moles HCl 0.00 moles NaOH moles NaOH = = moles NaOH m moles HCl = moles NaOH at equivalence pt: ph = (for SA/SB titration) d) 0.00 ml from part b, moles HCl = moles HCl 0.00 moles NaOH moles NaOH = 0.00 = moles NaOH m After neutralization: moles ecess base = moles moles = moles NaOH M OH moles = M NaOH = = M OH L poh = -log = 1.60 ph = = 1.98 Buffer, Titration and Solubility problems Key

4 . A 50.0 ml sample of 0.50 M HC H O acid is titrated with M NaOH for HC H O. Calculate the ph of the solution after the following volumes of NaOH have been added: a) 0 ml; b) ml; c) ml. a) 0 of base; only a weak acid is initially present so [H [HA HC H O H - C H O I C - E [ H [ C H [ HC H O O [H 5 = = 0.50( ) = = ph = -log =.5 b) of NaOH are added from part b, moles HC H O = moles HC H O moles NaOH moles NaOH = = moles NaOH m neutralization: HC H O OH - - C H O H O I C Final only acetate remains a weak base: moles [C H O = = M L - base hydrolysis: C H O H O HC H O OH - I C - E K b for C H O - = = [ HCH O [ OH K b = [ C H O = [OH - 10 = 0.115( ) = = poh = -log = 5.10 ph = = 8.90 At the equivalence point for a WA/SB titration, the ph > 7 due to the OH - produced by the conjugate base hydrolysis reaction. c) ml of NaOH are added from part b, moles HC H O = moles HC H O Buffer, Titration and Solubility problems Key 4

5 0.150 moles NaOH moles NaOH = = moles NaOH m moles ecess base = moles moles = moles NaOH M OH moles = M NaOH = = M OH - 0.L poh = -log =.06 ph = = *Ecess NaOH remains - this is the primary source of OH -. We can neglect the hydrolysis of the conjugate base because this would contribute a relatively small amount of OH - compared to the amount that comes directly from the ecess NaOH. Buffer, Titration and Solubility problems Key 5

6 SOLUBILITY PROBLEMS KEY 1. At 5 C, g of Ag CO dissolves in 1.0 L of solution. Calculate K sp for this salt g Ag CO 1mol Ag CO solubility = 1.0 L 75.8 g Ag CO = M Ag CO Ag CO (s) Ag (aq) CO (aq) K sp = [Ag [CO I 0 0 C E = molar solubility of Ag CO = M [CO = = M [Ag = = ( M) = M K sp = [ [ = Silver phosphate, Ag PO 4, is an insoluble salt that has a K sp = a) Calculate the molar solubility of Ag PO 4 in pure water. Ag PO 4 (s) Ag (aq) PO - 4 (aq) K sp = [Ag [PO - 4 I 0 0 C E K sp = () = = = M = molar solubility of Ag PO 4 in pure water b) Calculate the molar solubility of Ag PO 4 in a solution containing 0.00 M Na PO 4 (a soluble salt). soluble salt: Na PO 4 Na - PO 4 Phosphate is the common ion: [PO - 4 = [Na PO 4 = 0.00 M (since 1 mol Na PO 4 forms 1 mol PO - 4 ions) Ag PO 4 (s) Ag (aq) PO - 4 (aq) I C E 0.00 K sp = [Ag [PO = = () = 7 = = M = molar solubility of Ag PO 4 with a common ion Adding common ion decreases the solubility of Ag PO 4 Buffer, Titration and Solubility problems Key 6

7 . Does AgCl precipitate from a solution containing M Cl - and M Ag? K sp = Calculate Q for AgCl(s) Ag Cl - Q = [Ag [Cl - = [ [ = > ; Q > K sp Equilibrium shifts left & solid forms; AgCl precipitates 4. If you mi 10.0 of M Pb(NO ) with 5.0 of M HCl, does PbCl precipitate? K sp of PbCl = Pb(NO ) (aq) HCl(aq) PbCl (s) HNO (aq) Net ionic: Pb Cl - PbCl (s) Solubility reaction: PbCl (s) Pb Cl - Calculate Q for PbCl : Q = [Pb [Cl - [Pb = M Pb 10.0 = M Pb [Cl - = M Cl Q = ( )( ) = Q < K sp, so PbCl does not precipitate. = M Cl - Buffer, Titration and Solubility problems Key 7