Chpter 15 : xx, 4, 8, 0, 8, 44, 48, 50, 51, 55, 76, 8, 90, 96(90), 1(111) xx. NH 4 OH NH H O NH H O NH 4 H O 4.. HONH (q) H O(l) OH (q) HONH (q) HONH H O OH HONH initil 0.100 M ~0 0 x x x equilibrium 0.100 x x x 8 [OH ][HONH ] (x)(x) x b = 1.1 x 10 = = = [OHNH ] (0.100 x) 0.100 x If you ssume tht x << 0.100, then 0.100 x 0.100 x x 0.100 x 0.100 1.1 x 10 8 x = 1.1 x 10 9 x =. x 10 5 M = [OH ] (ssumption good) poh = log[oh ] = 4.48 ph = 14.00 poh = 9.5 b. HONH Cl(q) HONH (q) Cl (q) (Cl conj. bse SA) HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.100 M 0 ~0 x x x equilibrium 0.100 x x x x b = w for conjugte cid/bse b for HONH is 1.1 x 10 8. 1.0 x 10 [H O ][HONH ] (x)(x) x 9.1 x 10 (0.100 x) 0.100 x 14 7 = = = = = 8 1.1 x 10 [HONH ]
If you ssume tht x << 0.100, then 0.100 x 0.100 x x 0.100 x 0.100 9.1 x 10 7 x = 9.1 x 10 8 x =.0 x 10 4 M = [H O ] (ssumption good) ph = log[h O ] =.5 c. Pure H O ph = 7.00 d. Using either equilibrium: HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.100 M 0.100 ~0 x x x equilibrium 0.100 x 0.100 x x [H O ][HONH ] 14 1.0 x 10 7 (x)(0.100 x) = = = = 8 1.1 x 10 [HONH ] (0.100 x) 9.1 x 10 If you ssume tht x << 0.100, then 0.100 x 0.100 x (0.100 x) x (0.100) 0.100 x 0.100 9.1 x 10 7 x = 9.1 x 10 7 M (ssumption good) ph = log[h O ] = 6.04 8.. The dded HCl (s H O ) rects with HONH, forming HONH (s HONH Cl): HONH (q) H O (q) HONH (q) H O(l) HONH H O HONH H O I 0.100 mole/l x = 0 0.100 mole 0.00 mole F 0.100 0.00 = 0.080 mole 0 0.00 mole
0.080 mole HONH [ HONH ] = = 0.080 M HONH [HONH ] 0.00 mole HONH = = 0.00 M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.00 M 0.080 ~0 x x x equilibrium 0.00 x 0.080 x x [H O ][HONH ] 14 1.0 x 10 7 (x)(0.080 x) = = = = 8 1.1 x 10 [HONH ] (0.00 x) 9.1 x 10 If you ssume tht x is smll: x (0.080 x) x (0.080) 0.00 x 0.00 9.1 x 10 7 x =. x 10 7 M (ssumption good) ph = log[h O ] = 6.64 b. This is solution of SA nd WA. The H O from the WA is going to be negligible: [H O ] = 0.00 mole/ = 0.00 M ph = log[h O ] = 1.70 c. Sme s b. (even more so becuse H O is n even weker cid then HB ). ph = log[h O ] = 1.70 d. The dded HCl (s H O ) rects with just the HONH, forming HONH (s HONH Cl). This dds to the conjugte cid nd tkes wy from the bse:
HONH (q) H O (q) HONH (q) H O(l) HONH H O HONH I 0.100 mole/l x = 0.100 mole/l x 0.100 mole 0.00 mole = 0.100 mole F 0.100 0.00 = 0 0.100 0.00 = 0.080 mole 0.10 mole H O 0.080 mole HONH [ HONH ] = = 0.080 M HONH [HONH ] 0.10 mole HONH = = 0.10 M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.10 M 0.080 ~0 x x x equilibrium 0.10 x 0.080 x x [H O ][HONH ] 14 1.0 x 10 7 (x)(0.080 x) = = = = 8 1.1 x 10 [HONH ] (0.10 x) 9.1 x 10 If you ssume tht x is smll: x (0.080 x) x (0.080) 0.10 x 0.10 9.1 x 10 7 x = 1.4 x 10 6 M (ssumption good) ph = log[h O ] = 5.86 0.. This is solution of SB nd WB. The OH from the WB is going to be negligible: [OH ] = 0.00 mole/ = 0.00 M poh = log[oh ] = 1.70; ph = 14.00 poh = 1.0 b. The dded NOH (s OH ) rects with HONH (s HONH Cl), forming HONH :
HONH (q) OH (q) HONH (q) H O(l) HONH OH HONH H O I 0.100 mole/l x = 0 0.100 mole 0.00 mole F 0.100 0.00 = 0.080 mole 0 0.00 mole 0.00 mole HONH [HONH ] = = 0.00 M HONH [HONH ] 0.080 mole NH 4 = = 0.080 M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.080 M 0.00 ~0 x x x equilibrium 0.080 x 0.00 x x [H O ][HONH ] 14 1.0 x 10 7 (x)(0.00 x) = = = = 8 1.1 x 10 [HONH ] (0.080 x) 9.1 x 10 If you ssume tht x is smll: x (0.00 x) x (0.00) 0.080 x 0.080 9.1 x 10 7 x =.6 x 10 6 M (ssumption good) ph = log[h O ] = 5.44 c. Sme s. (even more so becuse H O is n even weker bse then B). poh = log[oh ] = 1.70; ph = 14.00 poh = 1.0 d. The dded NOH (s OH ) rects with HONH (s HONH Cl), forming HONH. This dds to the conjugte bse nd tkes wy from the cid:
HONH (q) OH (q) HONH (q) H O(l) HONH OH HONH H O I 0.100 mole/l x = 0.100 mole/l x 0.100 mole 0.00 mole = 0.100 mole F 0.100 0.00 = 0.080 mole 0 0.100 0.00 = 0.10 mole 0.10 mole HONH [HONH ] = = 0.10 M HONH [HONH ] 0.080 mole NH 4 = = 0.080 M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil 0.080 M 0.10 ~0 x x x equilibrium 0.080 x 0.10 x x [H O ][HONH ] 14 1.0 x 10 7 (x)(0.10 x) = = = = 8 1.1 x 10 [HONH ] (0.080 x) 9.1 x 10 If you ssume tht x is smll: x (0.10 x) x (0.10) 0.080 x 0.080 9.1 x 10 7 8. [NH ] = 0.75 M x = 6.1 x 10 7 M (ssumption good) ph = log[h O ] = 6. NH 4 Cl(q) [NH 4 ] = NH 4 (q) Cl (q) 1 mole NH4Cl 1 mole NH4 50.0 g NH4Cl x x 5.49 g NH Cl 1 mole NH Cl 4 4 = 0.95 M
NH (q) H O(l) OH (q) NH 4 (q) NH H O OH NH 4 initil 0.75 M ~0 0.95 M x x x equilibrium 0.75 x x 0.95 x 5 [OH ][NH 4 ] (x)(0.95 x) b = 1.8 x 10 = = [NH ] (0.75 x) If you ssume tht x smll, then: x (0.95) 0.75 = 1.8 x 10 5 x = 1.4 x 10 5 M = [OH ] (ssumption good) poh = log[oh ] = 4.84 ph = 14.00 poh = 9.16 44.. ph 7.40 8 [HO ] = 10 = 10 =.98 x 10 M [HCO ] = [HO ] = 4. x 10 [H CO ] 7 [HCO ] 4. x 10 4. x 10 = 10.8 [H CO ] [H O ].98 x 10 7 7 = = 8 [H CO ] 1 = [HCO ] 10.8 = 0.09 ph 7.15 8 b. [ H O ] = 10 = 10 = 7.08 x 10 M [HPO ] = [H O ] = 6. x 10 4 8 [HPO 4 ] 8 8 4 = = 8 4 [HPO ] 6. x 10 6. x 10 = 0.876 [H PO ] [H O ] 7.08 x 10 [H PO ] 1 = [HPO ] 0.876 4 4 = 1.1
c. The best buffer is one with 1:1 rtio of the cid (or bse) to the conjugte bse (or conjugte cid). This will give ph close to p. The p of H PO 4 is.1, which is fr removed from 7.1 7.. There would be very little H PO 4 t tht bsic ph. 48.. SA CB of SA not buffer n cidic solution b. SA WA not buffer no conjugte bse produced. Acidic. c. SA CB of WA yes, buffer. Hlf of the F ion would rect with the H O from HNO to give HF. This results in mixture of HF ( WA) nd F (its conjugte bse) buffer. d. SA SB not buffer bsic solution (NOH in excess). 50. Henderson Hsselblch is convenient for this type of clcultion. [A ] ph = p log [HA] The rection is: C H O (q) HCl(g) HC H O (q) Cl (q). For ph to equl p, the rtio must be 1:1. There re 1.0 mole CHO x 1.0 L = 1.0 mole C H O L To convert hlf (0.50 mole) into HCHO would require 0.50 mole HCl. b. [A ] 4.00 = 4.74 l og [HA] [A ] log = 0.54 ; [HA] 10 [A ] log [HA] [A ] 0.54 = = 10 = 0.88 [HA] Since x moles of moles CHO HCHO re produced from x moles of HCl(g) (by removing x ), the eqution bove becomes: 1.0 x x = 0.88 ; x = 0.78 = mole HCl(g)
c. [A ] 5.00 = 4.74 log [HA] [A ] l og = 0.6; [HA] 10 [A ] log [HA] [A ] 0.6 = = 10 = 1.8 [HA] 1.0 x x = 1.8 ; x = 0.5 = mole HCl(g) 51. 55.. HC H O (q) H O(l) H O (q) C H O (q) HC H O H O H O C H O initil 0.00 M ~0 0 x x x equilibrium 0.00 x x x 5 [HO ][CHO ] (x)(x) x = 1.8 x 10 = = = [HC H O ] (0.00 x) 0.00 x If you ssume tht x << 0.00, then 0.00 x 0.00 x x 0.00 x 0.00 1.8 x 10 5 x =.6 x 10 6 x = 1.9 x 10 M = [H O ] (ssumption good) ph = log[h O ] =.7
b. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x 0.100 L = 0.100 mole/l x 0.0500 L = 0 0.000 mole 0.00500 mole F 0.000 0.00500 = 0.0150 mole 0 0.0050 mole 0.0150 mole HC H O [HC H O ] = = 0.100 M HC H O 0.150 L 0.00500 mole CHO [CHO ] = = 0.0 M CHO 0.150 L HC H O (q) H O(l) C H O (q) H O (q) HC H O H O C H O H O initil 0.100 M 0.0 ~0 x x x equilibrium 0.100 x 0.0 x x 5 [HO ][CHO ] (x)(0.0 x) = 1.8 x 10 = = [HC H O ] (0.100 x) ssuming tht x is smll; x (0.0 x) x (0.0) 0.100 x 0.100 1.8 x 10 5 x = 5.4 x 10 5 M (ssumption good) ph = log[h O ] = 4.7 c. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x 0.100 L = 0.100 mole/l x 0.100 L = 0 0.000 mole 0.0100 mole F 0.000 0.0100 = 0.0100 mole 0 0.0100 mole
0.0100 mole HC H O [HC H O ] = = 0.0500 M HC H O 0.00 L 0.0100 mole CHO [CHO ] = = 0.0500 M CHO 0.00 L HC H O (q) H O(l) C H O (q) H O (q) HC H O H O C H O H O initil 0.0500 M 0.0500 ~0 x x x equilibrium 0.0500 x 0.0500 x x 5 [HO ][CHO ] (x)(0.0500 x) = 1.8 x 10 = = [HC H O ] (0.0500 x) ssuming tht x is smll; x (0.0500 x) x (0.0500) 0.0500 x 0.0500 1.8 x 10 5 x = 1.8 x 10 5 M (ssumption good) ph = log[h O ] = 4.74 d. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x 0.100 L = 0.100 mole/l x 0.1500 L = 0 0.000 mole 0.0150 mole F 0.000 0.0150 = 0.00500 mole 0 0.0150 mole [HC H O ] 0.0050 mole HC H O = = 0.000 M HC H O 0.50 L 0.01500 mole C H O [C H O ] 0.0600 M C H O 0.50 L = = HC H O (q) H O(l) C H O (q) H O (q)
HC H O H O C H O H O initil 0.000 M 0.0600 ~0 x x x equilibrium 0.000 x 0.0600 x x 5 [HO ][CHO ] (x)(0.0600 x) = 1.8 x 10 = = [HC H O ] (0.000 x) ssuming tht x is smll; x (0.0600 x) x (0.0600) 0.000 x 0.000 1.8 x 10 5 x = 6.0 x 10 6 M (ssumption good) ph = log[h O ] = 5. d. The dded OH (s OH ) rects with HC H O, forming C H O : HC H O (q) OH (q) C H O (q) H O(l) HC H O OH C H O I 0.00 mole/l x 0.100 L = 0.100 mole/l x 0.00 L = 0 0.000 mole 0.000 mole F 0.000 0.000 = 0.0 mole 0 0.000 mole 0.000 mole C H O [C H O ] 0.0667 M C H O 0.00 L = = C H O (q) H O(l) HC H O (q) OH (q) C H O H O HC H O OH initil 0.0667 M 0 ~0 x x x equilibrium 0.0667 x x x [OH ][HC H O ] x b = = 5.6 x 10 = = (0.0667 x) w 10 [CHO ] x x ssuming tht x is smll; 5.6 x 10 0.0667 x 0.0667 x =.7 x 10 11 ; x = 6.1 x 10 6 M (ssumption good) poh = log[oh ] = 5.1; ph = 14.00 poh = 8.79 10
f. This is 0.100 mole/l x 0.50 L = 0.050 mole OH excess 0.050 mole OH 0.000 mole OH rected = 0.00500 mole OH [ OH ] 0.00500 mole OH = = 0.50 L 0.014 M OH poh = log[oh ] = 1.85; ph = 14.00 poh = 1.15 76.. Ag CO (s) Ag (q) CO (q) b. Ce(IO ) (s) Ce (q) IO (q) c. BF (s) B (q) F (q) 8.. PbI (s) Pb (q) I (q) PbI Pb I initil 0 0 x x equilibrium x x sp = [Pb ][I ] = 1.4 x 10 8 = x(x) = 4x x = 1.5 x 10 M = [Pb ] = [PbI ] = molr solubility b. CdCO (s) Cd (q) CO (q) CdCO Cd CO initil 0 0 x x equilibrium x x sp = [Cd ][CO ] = 5. x 10 1 = x x =. x 10 6 M = [Cd ] = [CO ] = [CdCO ] = molr solubility c. Sr (PO 4 ) (s) Sr (q) PO 4 (q) Sr (PO 4 ) Sr PO 4 initil 0 0 x x equilibrium x x sp = [Sr ] [PO 4 ] = 1 x 10 1 = (x) (x) = 108x 5 x = x 10 7 M = [Sr (PO 4 ) ] = molr solubility
90.. Ag SO 4 (s) Ag (q) SO 4 (q) Ag SO 4 Ag SO 4 initil 0.10 M 0 x x equilibrium 0.10 x x sp = [Ag ] [SO 4 ] = 1. x 10 5 = (x) x = 4x x = 1.4 x 10 M = [SO 4 ] = molr solubility b. AgNO (q) Ag (q) NO (q) (ionizes completely) [Ag ] = [AgNO ] = 0.10 M Ag SO 4 (s) Ag (q) SO 4 (q) Ag SO 4 Ag SO 4 initil 0.10 M 0 x x equilibrium 0.10 x x sp = [Ag ] [SO 4 ] = 1. x 10 5 = (0.10 x) x Assume x smll: 1. x 10 5 (0.10) x x = 1. x 10 M (ssumption good) = [SO 4 ] = molr solubility c. SO 4 (q) (q) SO 4 (q) (ionizes completely) [SO 4 ] = [ SO 4 ] = 0.0 M Ag SO 4 (s) Ag (q) SO 4 (q) Ag SO 4 Ag SO 4 initil 0 0.0 M x x equilibrium x 0.0 x sp = [Ag ] [SO 4 ] = 1. x 10 5 = (x) (0.0 x) Assume x smll: 1. x 10 5 4x (0.0) x =.9 x 10 M (ssumption good) = [SO 4 ] = molr solubility
90.. AgF, becuse F is the CB of WA b. Pb(OH), becuse OH is SB c. Sr(NO ), becuse NO is the CB of WA d. Ni(CN), becuse CN is the CB of WA 111.. HC H O OH C H O H O 1 =? The reverse is b of the CB of WA: C H O H O HC H O OH 1.0 x 10 = = = 5.6 x 10 10 14 w b 5 1.8 x 10 1 1 = = = 1.8 x 10 9 1 10 b 5.6 x 10 b. C H O H O HC H O H O 1 =? The reverse is of WA: HC H O H O C H O H O 1 1 = = = 5.6 x 10 4 1 5 1.8 x 10 c. This rection is just: H O OH H O 1 =? This is the reverse of the w rection: H O H O OH w = 1.0 x 10 14 1 1 = = = 1.0 x 10 14 1 14 w 1.0 x 10