Review Guide, Sample Problems-- now with solutions
- Assume that 1.0 % of acetic acid vapor dimerizes at 2.5 atm pressures.
2 CH3CO2H (g) ---> {CH3CO2H}2 (g)
Evaluate Keq
-
we start with P(CH3CO2H) = 2.5 atm
- we lose 1% by reactions (2.475 atm remains)
- this 0.025 atm reacts,but becomes (1/2)(0.025 atm) of the dimer
- K = P(dimer) / P(acid)2
- K p= (0.0125) / (2.475)2
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A 0.050 M solution of a weak acid has a pH of 4.55. What is Ka for the acid?
-
[H3O+] = 10-4.55= 2.82 x 10-5
- [anion] = same
- [acid] = 0.050 - 2.82 x 10-5 which is nearly the same as 0.050
- ka = (2.82 x 10-5)2 / 0.050 = ..
.
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The salt QF3 --> Q3+ (aq) + 3F- (ag) and has a solubility of 2.1 x 10-4 moles per liter.
Evaluate Ksp
-
typo, may have read QMF3 by mistake, sorry)
- [Q3+] =2.1 x 10-4
- [F-] = 3 x 2.1 x 10-4 =6.3 x 10-4
- Ksp = (2.1 x 10-4) * (6.3 x 10-4)3
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Ksp for Ba(OH)2 is 3 x 10-3. If we mix equal volumes of 0.05M BaCl2 and 0.020 M NaOH, will we get a precipitate to form?
-
remember, simply mixing solutions dilutes each
- start with 0.025 M Ba2+ and 0.010 M OH-
- Q = [Ba2+] [OH-]2 = 0.025 * (0.010)2 = 2.5 x 10-6
- Q << Ksp so no precipitate
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NH4+ (or NH4Cl ) acts like a weak acid
Kb for ammonia is 1.75 x 10-5
what is Ka for ammonium ion?
-
Ka * Kb = Kw
- Ka = 1.0 x 10-14 / 1.75 x10-5 = 5.71 x 10-10
what is the pH of a 0.10 M solution of ammonium chloride
-
NH4+ + H2O ---> H3O+ + NH3
- [H3O+] = x = [NH3]
- [NH4+] = 0.10 -x
- x2/ (0.1-x) = 5.71 x 10-10
- can ignore x in the 0.1-x term
- x = sqrt( 5.71 x 10-11) = 7.56 x 10-6
- pH = -log (ans) = 5.12
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What is the pH of a solution that is 0.10 M Acetic acid; Ka = 1.8 x 10-5
-
really the same as the NH4+ problem above, with a different Ka
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Kc = 138 for the reaction Fe3+ + SCN- ----> Fe(SCN)2+
If we start with 2.0 x 10-3M Fe3+ and 0.050 M SCN-, how much Fe(SCN)2+ forms?
-
let x = [Fe(SCN)2+] at equilibrium (also change in the amount)
- [Fe3+ ] = 0.0020 -x
- [SCN-] = 0.0500 -x
- at equil K= 138 = x / {(0.0020-x)*(0.050-x)}
- this is a quadratic
- but x can't exceed 0.002 (all Fe3+ used) so [SCN-] can't be less than 0.048
- approximate 138 = x /(0.0020-x)*(0.050 ___)
- this is easy to solve for X
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What is the pH of a solution that is 0.030M Acidic acid and 0.020 M potassium acetate?
Ka for acetic acid is 1.8 x 10-5.
-
pka = - log(1.8 x 10-5) = 4.75
- pH = pKa + log [(Ac)/(HAc)] = 4.75 + log( .003/0.002)
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Concepts--
- Why is sodium oxalate more soluble in 0.1 M HCl than it is in water?
-
oxalate ion reacts with H+ forming the waek acid
- therefore, reduces oxalte ion concentration and more salt dissolves
.
- Why is PbCl2 less soluble in 0.1M HCl than it is in water?
-
Cl- is a common ion
- Cl- is not accociated with a weak acid, so it ignores the H+
- Will adding a drop of HCl (1M) to 5 ml of 1 M solution of acetic acid
- a.
raise the pH significantly
(The HCl alone now fixes the pH)
b. lower the pH significantly
- c, have relatively little effect on the pH
- Will adding a drop of 1M Acetic acid to 5 ml of 1 M HCl
- (same three choices)
-
(c.-- no change, the HCl fixes the H+
- If you add 1 drop (1/20 ml) of 1 M NaOH to 5.0 ml of water, what will be the pH?
-
[OH-] = 1.0 M x (0.05/5) where the ( ) is the dilution ratio
- [OH-] = 0.010
- pOH = 2
- pH = 14 - pOH = 14.0-2.0 = 12.00
- If you added that NaOH to a solution that is 0.1M Acetic acid and 0.10 Sodium acetate,
- why would the pH change be very small
-
this is a buffer mixture
- the OH- is used up converting Acetic acid to acetate ion
- the pH still stays close to pKa (see an ealier problem)
- Adding sodium thiosulfate to water forms a solution that is several degrees cooler than the starting water.
- Is sodium thiosulfate more soluble in hot water or in cold water?
-
the reaction (dissolving) is endothermic
- Le Chatlier says equil constant is increased if temp rises
- so solubility is greaster in hot water.
- If you slowly add 0.1M NaOH to a solution of HCl and acetic acid could you...
- produce a solution that is mainly NaCl and Acetic acid?
-
yes.. the NaOH removes H3O+ (HCl) but won't attack weak acid ntil the HCl is gone.
- produce a solution that is mainly HCl and sodium acetate?
-
no (see above)
H3O+ + acetate ---> acetic acid
- a solution that has lost half of the HCl and half of the acetic acid?
- a solution where OH- and acetate ion both exist at modest concentrations?
-
sure, the OH- will eventually convert the acetic acid t acetate ion
- just keep adding more OH- until it matches
- a solution that is pH 7.00 ?
-
sure, we start acidic (pH 1.0) and gradually increase pH
- at some point we will pass through pH 7.00
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- The freezing point of a 1.0 M solution of glucose (a sugar) is -1.86 oC
- The freezing point of a 1.0 M solution of a weak acid is 2.01 oC
- why the difference?
-
since the acid partly reacts, the total concentration of ions > 1.0 M
- here it acts like a (2.01/1.860 = 1.081molal
- so we get 0.016 molal H3O+, 0.081 molal anion and we keep 0.919 M acid
- Ka = 0.081 x 0.081 / 0.919
- HA --> H+ + A-
- how could this information be used to find Ka for the acid
- ----------------
- CaCO 3 (solid limestone) can be heated in air to completely decompose to CaO + CO2 (g)
- If CaO is left exposed to normal air, it picks up CO2 and becomes CaCO3 again.
- can you explain and use this information to tell something about the reaction
-
basically, the equiibrium shifts with temperature
- at high temperature, we favor gas and CaO
- at lower temepratures, even the 0.01% CO2 in air will react.---------------
- so the reaction must be endothermic