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Algemene chemie oefeningen hoofdstuk 16
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Algemene chemie oefeningen Mastering Chemistry met uitwerking (Hoofdstuk 16)
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Chemie oefeningen hoofdstuk 161. Wat is de pH van een 0.150 M CH 2O.250 HCOONa buffer? De Ka van CH2O2 is 1.8 * 10-4
Oplossing:
[ H 3 O ] +¿[HCOONa] [CH 2 O 2] 0.150 M
Buffer dus Ka = ¿ dus H3O = Ka * = 1.8 * 10-4 * =
[CH 2 O 2] [ HCOONa] 0.250 M
-4
1.08 * 10
pH = - log (1.08 * 10-4) = 3.97
2. Wat is de verhouding [CH3CO2-]/[CH3CO2H] die nodig is om een buffer te maken met pH 4.14? Ka
= 1.8 * 10-5 voor CH3CO2H.
Oplossing:
[CH 3 CO 2 H ] H 3 O [CH 3 CO 2 H ]
Buffer dus H3O = Ka * dus =
[CH 3 CO 2] Ka [CH 3 CO 2]
Met H3O = 10-pH = 7.244 * 10-5
H 3O [CH 3 CO 2 H ] [CH 3 CO 2]
= 4.02 = dus = ¼ = 0.25:1
Ka [CH 3 CO 2] [CH 3 CO 2 H ]
3. Welke pH bekom je als je 0.0125 mol HCl toevoegt bij 0.125 ml van een 0.150M CH 2O2/0.100M
HCOONa? De Ka van CH2O2 is 1.8 * 10-4.
Oplossing:
Neutralisatie van een sterk zuur gaat 100%
H3O+ + HCOO- <-> HCOOH + H2O
Voor reactie n (HCOOH) = 0.150 mol/l * 0.150 l = 0.0225 mol
n (HCOO-) = 0.100 M * 0.150 l = 0.015 mol
na reactie: n (HCOOH) = 0.0225 mol + 0.0125 mol = 0.0350 mol
n (HCOO-) = 0.0025 mol
Ka∗[HCOOH ] = (0.2333 M * 1.8*10-4) / (0.01667 M) = 0.00252 M
Ka = ¿ ¿ H3O =
¿¿
CHCOOH = n/V = 0.0350 mol / 0.150 l = 0.2333 M
CHCOO^- = n/V = 0.0025 mol / 0.150 l = 0.016667 M
pH = - log H3O = 2.60
, 4. Wat is de pH voor de oplossing die gevormd wordt wanneer 25 ml van 0.173M NaOH wordt
toegevoegd aan 35 ml van 0.342 M HCL?
Oplossing:
NaOH + HCL <-> H2O + NaCl
n=C*V=0.025*0.173 n = 0.035*0.342
= 0.004325 mol = 0.01197 mol 0.01197 mol
Blijft over: 0.01197 – 0.004325
= 0.007645 mol -> C = n/V = 0.007645 / (0.025 + 0.035) = 0.127 M
pH = - log (zuur) = - log (0.127) = 0.89
5. Welk volume van 5*10-3M HNO3 is nodig om 100 ml van 5*10-3 M Ca(OH)2 te titreren tot het
equivalentiepunt?
Oplossing:
Ca(OH)2 + 2 HNO3 <-> 2 H2O + Ca(NO3)2
1 mol 2 mol
Dus het volume voor HNO3 moet dubbel zijn , dus 200 ml
6. Wat is de pH van de oplossing die je maakt door 40 ml van 0.10 CH 3COOH toe te voegen aan 0.10
M NaOH? Ka = 1.8 * 10-5 voor CH3COOH
Oplossing:
CH3COOH + NaOH <-> H2O + CH3COONa
1 mol 1 mol 1 mol
n = C*V=0.1*0.04 n = 0.1*0.01
= 0.004 mol = 0.001 mol = 0.001 mol
Dus 0.003 mol over!
M CH3COOH = n / V = 0..05 = 0.06 M
M CH3COONa = n / V = 0..05 = 0.02 M
pH via Henderson-Hasselbach want buffer: pH = pKa + log (zout/zuur) = 4.2676
7. Wat is de pH van een oplossing die gevormd wordt door 10 ml 0.1 M CH 3COOH met 10 ml 0.1 M
KOH? Ka= 1.8 * 10-5
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