# Hydroxylamine, HONH2, Is a Weak Base EXAM 2 KEY

PART A: EQUILIBRIA

1) Consider the Following OVERALL Reaction (15 pts)

2NO + 2CO < == > 2NCO + O2
In which the Rate of the forward reaction is equal to Rf = kf [NO][CO]2
and the Rate of the reverse (backward) reaction is equal to Rb = kb [NCO]2 [O2] /[ NO]

Under what conditions would the above reaction be defined as at Equilibrium? (1 pts)

When the Forward Rate = Backward Rate ; Rf = Rb

What defines the Equilibrium Constant Keq in terms of the Rate Laws(1 pts)

Keq= kf / kb

What defines the Equilibrium Constant Keq in terms of CONCENTRATIONS(1 pts)

Keq = [NCO]2 [O2]/ ( [NO] 2[CO]2)

If AT EQUILIBRIUM, one finds that the concentrations of NCO and O2 are 0.135M,

while the concentrations of NO and CO are 0.0214 M and 0.107 M respectively;

What would be the NUMERIC VALUE of the Equilibrium Constant? (3 pts)

Keq= =0.135^2*0.135/(0.0214^2*0.107^2)=469.25

Staring with the concentrations as defined above (d); if one removed 0.01 moles/L of BOTH NO and O2 from the reaction; How would that change the Reaction Quotient Q. NUMERICALLY and what would THEN happen to Q?

(5 pts)

=0.135^2*0.125/(0.0114^2*0.107^2) / 1531.088

Q would now be larger than Keq and thus the reaction would need to move to the LEFT back to the reactants

Now in a more QUANTITAVE manner, using the Keq, show how one would solve the above problem to obtain the final equilibrium concentration of [A]. SETUP ONLY(5 pts)

=(0.135-2x)^2*(0.125-x)/{ (0.0114+2x)^2* (0.107+2x)^2) } =Keq = 469.25

PART B: Le Chatelier's principle (10 pts)
Consider the Reaction: 3H2 (gas) + N2(gas) < == > 2 NH3(gas)
is found to be EXOTHERMIC and have a Kpeq = 3.70 x 10-3 atmospheres-2

a) Discuss three ways that one can use Le Chatelier's principle (to include Reactants/Products, Temperature, Volume and Pressure) to produce the maximum quantity of NH3. (3pts)

1) Increase Reactants and/or Remove Product

2) INCREASE PRESSURE or REDUCE VOLUME of system which will push reaction to right in order to reduce pressure(fewer moles on right

3) Because EXOTHERMIC; reduce the temperature which will pull reaction to the right

b) If one began with only H2 and N2 gas both at 0.21 atm partial pressures; what would be the final pressure of NH3. SETUP only (5pts)

3H2 (gas) + N2(gas) < == > 2 NH3(gas) Kpeq=3.7 x 10-3 =(2x)2 /{ (0.21-3x)3*(0.21-x) } = > xexact= 0.0013

then [NH3] = 2x

c) Assume now that the one guesses the change x to be 0.006 in H2 and N2 in achieving equilibria; utilizing this value, is the Q too big or too small in it’s effect and how should x be changed?(2pts)

Q=9.97E-02 which is TOO BIG Thus too much Product and thus X is TOO BIG (Reduce x)

PART C ( Titration) (5pts)

Draw a reasonable titration curve for the reaction of 100 ml 0.2M Pyruvic acid (pKa= 2.39) with 0.1M Potassium hydroxide (assume totally soluble). Contrast this with the titration curve for a 0.2M Nitric acid

Note the THREE(3) important pH’s that help define the shape of this graph

GRAPH TO INCLUDE S shaped surve for weak acid with 3 key pH points at 0, 100 200mls of KOH

=0.5*(2.39-LOG(0.2))1.54

= Pyruvic Acid=1.54

HCl= 0.70

pH= pKa –log( [HA]/[A])=pKa2.39 (1/2 EQUIVALENCE)

pH(salt)= 7+ ½ (pKa+log[A-]=7+0.5*(2.39+LOG((0.02/0.3)))7.61 (REMEMBER DILUTION for A)

PART D ( Acid-Base Chemistry ) (10x5=50 pts)

1) Using the fact that pKa(HCOOH) = 3.75; show at least TWO ways to construct a BUFFER solution at pH=3.75 using the following:

0.15M solution of Sodium Hydroxide
0.15M solution of Sodium Formate
0.15M solution of Sodium Fluoride
0.15M solution of Hydrochloric Acid
0.15M solution of Hydrofluoric Acid
0.15M solution of Sodium Chloride
0.15M solution of Formic Acid

1) Equal Volumes both HFm and NaFm

2) One Volume of HFm and ½ that volume of NaOH making Fm-

3) One Volume of NaFm and ½ that volume of HCl to make HFm

2) Which of the following acids, if in solutions of equal concentration, is the least acidic? (2pt) most acidic

a) 0.1M propanoic acid pKa=4.86b) 0.1M octanoic acid pKa=4.89

c) 0.1M uric acid, pKa=3.89 d) 0.1M formic acid pKa=3.75

e) All of these acids are equally acidic because they are all of equal concentration.

0.1M octanoic acid pKa=4.89 least

0.1M formic acid pKa=3.75 most

3) What is the pH of an aqueous solution of 0.574 M 2,3 dichloro-pyridine pKb= 6.89

(a weak base with the formula C5Cl2H5N)

BASE: pOH= ½(Kb-log(0.574) =0.5*(6.89-LOG(0.574))3.57 pH=10.43

4) How many milligrams of Magnesium hydroxide(Molar Mass=58.305) would be need to be added to one liter of solution (assume STRONG and it all dissolves) to give an aqueous solution with a pH of 10.73

pOH= 14-10.733.27

[OH] = =10^-3.270.000537032

but [Mg(OH)2 ] = 0.5[OH]=2.69E-04

1.57E-02g  15.7 mg

5) If the pH of a solution of NaF is adjusted with a strong acid or base. Using the fact that pKa(HF) = 3.17; when is half of the total fluoride in the form of Fluoride ION and when is 90% of the total fluoride in the form of F-?

pH= pKa - log (0.5/0.5) = pKa = 3.17

pH = pKa - log(0.10/0.90) = 4.12

6) Which of the following aqueous solutions are good buffer systems AND WHY!!?

0.24 M hydrochloric acid + 0.22 M potassium chloride / NO(strong acid)
0.31 M ammonia + 0.34 M amonium chloride / YES
0.04 M calcium fluoride + 0.47 M hydrofluoric acid / NO(could be BUT KF too small
0.31 M sulfuric acid + 1.24 M methylamine / YES(Half of the base will be converted to conjugate acid
0.36 M potassium sulfate + 0.26 M barium sulfate / NO(both salts)
0.45 M hydrocyanic acid + 0.39 M sodium cyanide / YES

7) What is the pH of a solution contains 0.202 M methylamine iodide and 0.389 M methylamine. (pKb(CH3NH2)= 3.37)
pH of BASE BUFFER: pOH= pKb - log ([B]/[BH+]) = 3.37 - log(0.389/0.202) = 3.09  pH= 10.91

8) 100ml of 0.481 M nitrous acid (pKa = 3.35) ; 100 ml of 0.314 M sodium nitrite and 100 ml of 0.3M sodium hydroxide are mixed together , what is the pH of the resulting solution? (Assume that the volume are additive)

moles of initial acid / 0.0481
moles of initial base / 0.03
moles of initial nitrite / 0.0314
moles of final acid / 0.0181
moles of final nitrite / 0.0614
pH=pKa-log(acid/conj base) / 3.88

9) What volume of a 0.5 M nitric acid solution is required to titrate to equivalence 26.0 mL of a 0.125 M ammonium hydroxide (pKb= 4.75) solution ? And what is the resulting pH

MV=MV
(0.5)*x = 0.125*0.026 / 0.0065 / L / 6.5 ml
Strong Acid/WEAK base ==> No excess Acid/Base and BUT contains Salt of Weak Base (Acidic)
Molarity of NH4+ = moles of NH4/(Total Volume) = (0.026*0.125)/(0.0065+0.026) = 0.100 M
pH= 7- ½ *(4.75+log(0.10)) = 5.125

10) Write a net ionic equation to show that Thioacetamide (H2C2SNH2) behaves as a

Bronsted-Lowry base in WATER. (2pt)

BL base BL acid BL acid BL base

H3C2SNH2 + H2O < == > H3C2SNH3+OH-

Part E. Consider the following Titration Conditions (8x5=40pts)

:

0.20 M ACID / 0.30 M BASE / Equation/work / pH
50 ml Chloroacetic
Acid (pKa=2.85) / 15 ml Ca(OH)2 / moles of acid = MaVa / 0.010
moles of base = MbVb / 0.009
Total Volume= / 0.065 / ml
Weak Acid (HH Eq) pH= / 3.80
50 ml HCl / 50 ml NH3
(pKb = 4.75) / moles of acid = MaVa / 0.010
moles of base = MbVb / 0.015
Total Volume= / 0.100 / ml
WEAK Base (HH Eq) / pH= / 8.95
120 ml Nitrous Acid
(pKa)=3.35 / 40 ml NaOH / At Half Equivalence Point
pH= pKa-log([HA]/[A-]) but log(1)=0
thus pH= 3.35
40ml of Sulfuric Acid / 40 ml of NaOH / moles of acid = MaVa / 0.016
moles of base = MbVb / 0.012
Total Volume= / 0.080 / ml
STRONG Acid Won / pH= / 1.30
30 ml Formic Acid
(pKa)=3.17 / 10 ml of Ca(OH)2 / moles of acid = MaVa / 0.006
moles of base = MbVb / 0.006
Total Volume= / 0.040 / ml
If Salt(WA/SB) / 8.17
20 ml of Cyanic Acid (HCNO)
pKa = 3.46 / No Base / moles of acid = MaVa / 0.004
moles of base = MbVb / 0.000
Total Volume= / 0.020 / ml
pH(WA alone)= / 2.08
No Acid / 25 ml of Methylamine pKb=3.34 / moles of acid = MaVa / 0.000
moles of base = MbVb / 0.008
Total Volume= / 0.025 / ml
pH(WB alone)= / 12.07
37.5 ml of HCl / 25 ml of Methylamine pKb=3.34 / moles of acid = MaVa / 0.0075
moles of base = MbVb / 0.0075
Total Volume= / 0.063 / ml
Salt (WB/SA) / 5.79