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 = 1.810-5 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 0.0 0.15 0 C - E 0.0-0.15 K b = [ NH 4 [ NH [ OH Approimation: ignore, terms: 1.810-5 = ( 0. 15 ) ( 0. 0) = [OH - =.610-5 M poh = -log.610-5 = 4.44 ph = 14 4.44 = 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 =1.810-4 a) What is the ph of this buffer solution? Salt: NaCHO Na - CHO Acid ionization rn: HCHO H - CHO I 0.0 0 0.0 C - E 0.0-0.0 Approimation: ignore, terms [ H [ CHO [ HCHO 1.810-4 = ( 0.0) ( 0.0) = [H = 1.10-4 ph = -log[h = -log 1.10-4 =.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
neutralization reaction: OH - - HCHO CHO Initial 0.09 0.14 0.1 Change -0.09-0.09 0.09 Final 0 0.11 0.4 Acid ionization rn: HCHO H - CHO I 0.11 0 0.4 C - E 0.11-0.4 [ H [ CHO [ HCHO 1.810-4 = ( 0.4) ( 0.11) H O = [H = 8.510-5 ph = -log[h = -log 8.510-4 = 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 0.09 0.1 0.14 Change -0.09-0.09 0.09 Final 0 0.18 0.17 Acid ionization rn: HCHO H - CHO I 0.17 0 0.18 C - E 0.17-0.18 [ H [ CHO [ HCHO 1.810-4 = ( 0.18) ( 0.17) (175) = 0.09 M = [H = 1.710-4 ph = -log[h = -log 1.710-4 =.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
TITRATION PROBLEMS KEY 1. A 0.00 sample of 0.150 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) 10.00 ml; c) 15.0 ml; d) 0.00 ml. a) 0 of NaOH added only SA is present initially: For strong acid: [H = [HCl = 0.150 M HCl ph = -log[h = -log(0.150) = 0.84 b) 10.00 of NaOH neutralization reaction: HCl NaOH NaCl H O SA SB 0.150 moles HCl moles HCl = 0.00 =.0010 - moles HCl m 0.00 moles NaOH moles NaOH = 10.00 =.0010 - moles NaOH m After neutralization: moles ecess acid =.0010 - moles -.0010 - moles = 1.0010 - moles HCl M H 1.0010 moles = M HCl = = 0.0 M 0.0000 L ph = - log [H = - log 0.0 = 1.478 c) 15.0 ml of NaOH From part b, moles HCl =.0010 - moles HCl 0.00 moles NaOH moles NaOH = 15.00 =.0010 - moles NaOH m moles HCl = moles NaOH at equivalence pt: ph = 7.000 (for SA/SB titration) d) 0.00 ml from part b, moles HCl =.0010 - moles HCl 0.00 moles NaOH moles NaOH = 0.00 = 4.0010 - moles NaOH m After neutralization: moles ecess base = 4.0010 - moles.0010 - moles = 1.0010 - moles NaOH M OH - 1.0010 moles = M NaOH = = 0.050 M OH - 0.040 L poh = -log 0.050 = 1.60 ph = 14 1.60 = 1.98 Buffer, Titration and Solubility problems Key
. A 50.0 ml sample of 0.50 M HC H O acid is titrated with 0.150 M NaOH. 1.810-5 for HC H O. Calculate the ph of the solution after the following volumes of NaOH have been added: a) 0 ml; b) 166.7 ml; c) 180.0 ml. a) 0 of base; only a weak acid is initially present so [H [HA HC H O H - C H O I 0.50 0 0 C - E 0.50- [ H [ C H [ HC H O O [H 5 = = 0.50(1.8 10 ) =.010-1.810-5 = ph = -log.010 - =.5 b) 166.7 of NaOH are added from part b, moles HC H O =.510 - moles HC H O 0.150 moles NaOH moles NaOH = 166.7 =.5010 - moles NaOH m neutralization: HC H O OH - - C H O H O I 0.05 0.050 0 C -0.05-0.05 0.05 Final 0 0 0.05 only acetate remains a weak base: -.5 10 moles [C H O = = 0.115 M 0.167 L - base hydrolysis: C H O H O HC H O OH - I 0.115 0 0 C - E 0.115-14 1 10 K b for C H O - = = 5.610-10 5 1.810 [ HCH O [ OH K b = [ C H O = [OH - 10 = 0.115( 5.6 10 ) = 8.010-6 0.50 5.610-10 = 0.115 poh = -log 8.010-6 = 5.10 ph = 14 5.10 = 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) 180.0 ml of NaOH are added from part b, moles HC H O =.510 - moles HC H O Buffer, Titration and Solubility problems Key 4
0.150 moles NaOH moles NaOH = 180.00 =.710 - moles NaOH m moles ecess base =.710 - moles -.510 - moles =.010 - moles NaOH M OH -.010 moles = M NaOH = = 8.710 - M OH - 0.L poh = -log 8.710 - =.06 ph = 14.06 = 11.94 *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
SOLUBILITY PROBLEMS KEY 1. At 5 C, 0.049 g of Ag CO dissolves in 1.0 L of solution. Calculate K sp for this salt. 0.049 g Ag CO 1mol Ag CO solubility = 1.0 L 75.8 g Ag CO = 1.10-4 M Ag CO Ag CO (s) Ag (aq) CO (aq) K sp = [Ag [CO I 0 0 C E = molar solubility of Ag CO = 1.10-4 M [CO = = 1.10-4 M [Ag = = (1.10-4 M) =.610-4 M K sp = [.610-4 [1.10-4 = 8.810-1. Silver phosphate, Ag PO 4, is an insoluble salt that has a K sp = 1. 10-0. 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 = () 1.10-0 = 7 4 4 = 4.810 - = 4.710-6 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 0 0.00 C E 0.00 K sp = [Ag [PO - 4 1.10-0 = = () 0.00 6.510-19 = 7 =.410-0 =.910-7 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
. Does AgCl precipitate from a solution containing 1.0 10-5 M Cl - and 1.5 10-4 M Ag? K sp = 1.8 10-10 Calculate Q for AgCl(s) Ag Cl - Q = [Ag [Cl - = [1.510-4 [1.010-5 = 1.510-9 1.510-9 > 1.810-10 ; Q > K sp Equilibrium shifts left & solid forms; AgCl precipitates 4. If you mi 10.0 of 0.0010 M Pb(NO ) with 5.0 of 0.015 M HCl, does PbCl precipitate? K sp of PbCl = 1.6 10-5 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 = 0.0010 M Pb 10.0 = 6.710-4 M Pb 10.0 5.0 [Cl - = 0.015 M Cl - 5.0 5.0 10.0 Q = (6.710-4 )(5.010 - ) = 1.710-8 Q < K sp, so PbCl does not precipitate. = 5.010 - M Cl - Buffer, Titration and Solubility problems Key 7