HONH 3 initial M - ~0 0 -x - +x +x equilibrium x - x x. ] (x)(x) x b. If you assume that x << 0.100, then x
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1 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 M ~0 0 x x x equilibrium x x x 8 [OH ][HONH ] (x)(x) x b = 1.1 x 10 = = = [OHNH ] (0.100 x) x If you ssume tht x << 0.100, then x x x x x 10 8 x = 1.1 x 10 9 x =. x 10 5 M = [OH ] (ssumption good) poh = log[oh ] = 4.48 ph = 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 M 0 ~0 x x x equilibrium x x x x b = w for conjugte cid/bse b for HONH is 1.1 x x 10 [H O ][HONH ] (x)(x) x 9.1 x 10 (0.100 x) x 14 7 = = = = = x 10 [HONH ]
2 If you ssume tht x << 0.100, then x x x x 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 M ~0 x x x equilibrium x x x [H O ][HONH ] x 10 7 (x)(0.100 x) = = = = x 10 [HONH ] (0.100 x) 9.1 x 10 If you ssume tht x << 0.100, then x x (0.100 x) x (0.100) x x 10 7 x = 9.1 x 10 7 M (ssumption good) ph = log[h O ] = 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 mole/l x = mole 0.00 mole F = mole mole
3 0.080 mole HONH [ HONH ] = = 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 x x x equilibrium 0.00 x x x [H O ][HONH ] x 10 7 (x)(0.080 x) = = = = 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 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:
4 HONH (q) H O (q) HONH (q) H O(l) HONH H O HONH I mole/l x = mole/l x mole 0.00 mole = mole F = = mole 0.10 mole H O mole HONH [ HONH ] = = 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 x x x equilibrium 0.10 x x x [H O ][HONH ] x 10 7 (x)(0.080 x) = = = = 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 x 10 7 x = 1.4 x 10 6 M (ssumption good) ph = log[h O ] = 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 = poh = 1.0 b. The dded NOH (s OH ) rects with HONH (s HONH Cl), forming HONH :
5 HONH (q) OH (q) HONH (q) H O(l) HONH OH HONH H O I mole/l x = mole 0.00 mole F = mole mole 0.00 mole HONH [HONH ] = = 0.00 M HONH [HONH ] mole NH 4 = = M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil M 0.00 ~0 x x x equilibrium x 0.00 x x [H O ][HONH ] x 10 7 (x)(0.00 x) = = = = x 10 [HONH ] (0.080 x) 9.1 x 10 If you ssume tht x is smll: x (0.00 x) x (0.00) x 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 = 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:
6 HONH (q) OH (q) HONH (q) H O(l) HONH OH HONH H O I mole/l x = mole/l x mole 0.00 mole = mole F = mole = 0.10 mole 0.10 mole HONH [HONH ] = = 0.10 M HONH [HONH ] mole NH 4 = = M HONH HONH (q) H O(l) HONH (q) H O (q) HONH H O HONH H O initil M 0.10 ~0 x x x equilibrium x 0.10 x x [H O ][HONH ] x 10 7 (x)(0.10 x) = = = = x 10 [HONH ] (0.080 x) 9.1 x 10 If you ssume tht x is smll: x (0.10 x) x (0.10) x x [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 NH g NH4Cl x x 5.49 g NH Cl 1 mole NH Cl 4 4 = 0.95 M
7 NH (q) H O(l) OH (q) NH 4 (q) NH H O OH NH 4 initil 0.75 M ~ 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 = poh = ph [HO ] = 10 = 10 =.98 x 10 M [HCO ] = [HO ] = 4. x 10 [H CO ] 7 [HCO ] 4. x x 10 = 10.8 [H CO ] [H O ].98 x = = 8 [H CO ] 1 = [HCO ] 10.8 = 0.09 ph b. [ H O ] = 10 = 10 = 7.08 x 10 M [HPO ] = [H O ] = 6. x [HPO 4 ] = = 8 4 [HPO ] 6. x x 10 = [H PO ] [H O ] 7.08 x 10 [H PO ] 1 = [HPO ] = 1.1
8 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 There would be very little H PO 4 t tht bsic ph 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)
9 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) 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 x 10 5 x =.6 x 10 6 x = 1.9 x 10 M = [H O ] (ssumption good) ph = log[h O ] =.7
10 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 L = mole/l x L = mole mole F = mole mole mole HC H O [HC H O ] = = M HC H O L mole CHO [CHO ] = = 0.0 M CHO 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 M 0.0 ~0 x x x equilibrium 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) x 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 L = mole/l x L = mole mole F = mole mole
11 mole HC H O [HC H O ] = = M HC H O 0.00 L mole CHO [CHO ] = = 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 M ~0 x x x equilibrium x x x 5 [HO ][CHO ] (x)( x) = 1.8 x 10 = = [HC H O ] ( x) ssuming tht x is smll; x ( x) x (0.0500) x 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 L = mole/l x L = mole mole F = mole mole [HC H O ] mole HC H O = = M HC H O 0.50 L mole C H O [C H O ] M C H O 0.50 L = = HC H O (q) H O(l) C H O (q) H O (q)
12 HC H O H O C H O H O initil M ~0 x x x equilibrium x x x 5 [HO ][CHO ] (x)( x) = 1.8 x 10 = = [HC H O ] (0.000 x) ssuming tht x is smll; x ( x) x (0.0600) x 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 L = mole/l x 0.00 L = mole mole F = 0.0 mole mole mole C H O [C H O ] 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 M 0 ~0 x x x equilibrium x x x [OH ][HC H O ] x b = = 5.6 x 10 = = ( x) w 10 [CHO ] x x ssuming tht x is smll; 5.6 x x x =.7 x ; x = 6.1 x 10 6 M (ssumption good) poh = log[oh ] = 5.1; ph = poh =
13 f. This is mole/l x 0.50 L = mole OH excess mole OH mole OH rected = mole OH [ OH ] mole OH = = 0.50 L M OH poh = log[oh ] = 1.85; ph = poh = 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
14 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 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 x (0.0) x =.9 x 10 M (ssumption good) = [SO 4 ] = molr solubility
15 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 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 w b x = = = 1.8 x 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 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 = = = 1.0 x w 1.0 x 10
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