CHEM 102 DWS #4 February 11, 1998
NAME(print) ,
(LAST) (FIRST)
1. A solution is prepared by taking 12.75 ml of a 6.00 M
hydrochloric acid solution, and diluting to a final volume
of 500 ml. What is the pH, pOH, [ H+1 ], and [ OH-1 ]?
2. A solution is prepared by weighing 3.582 g of benzoic acid,
C6H5COOH, and dissolving in enough water to yield a final
volume of 300 ml. What is the pH, pOH, [ H+1 ], and
[ OH-1 ]? (Ka benzoic acid = 6.6E-5)
3. What is the pH, pOH, [ H+1 ], and [ OH-1 ], for a 0.25 M
solution of potassium hydroxide?
4. What is the pH, pOH, [ H+1 ], and [ OH-1 ], for a 3.0 %
aqueous solution of ammonia? (Kb = 1.8E-5)
5. Indicate how the pH for the reaction would change:
HF (aq) = H+1(aq) + F-1(aq) if the changes shown
below were made:
a. adding some hydrochloric acid
b. adding some ammonia
c. adding some sodium fluoride
d. magnesium hydroxide
6. Suppose 10.0 g ammonium chloride are added to 500 ml of an
0.13 M aqueous solution of ammonia. What is the final pH,
pOH, [ H+1 ], and [ OH-1 ]?
7. The pH changes after 7.65 g sodium acetate are added to 350
ml of a 0.175 M solution of acetic acid. (Ka = 1.8E-5)
a. What is the pH of the original solution?
b. What is the pH of the solution after addition of sodium
acetate?
8. Suppose that 15.00 ml of 0.125 M NaOH are added to 75.00 ml
of 0.103 M formic acid, CHOOH. What is the final pH, pOH,
[ H+1 ], and [ OH-1 ]? (Ka formic acid = 2.1E-4)
9. Suppose that one Liter of 0.150 M propionic acid
(CH3CH2COOH, Ka = 1.4E-5)and one Liter of 0.125 M hydroiodic
acid, HI, are mixed. What is the pH of the mixed solutions?
ANSWERS
1. a. Conc. diluted solution: M = 12.75 * 6.0 / 500 = 0.153 M
b. HCl, strong acid, complete ionization (CI) (one way Rx):
HCl ---> H+1 + Cl-1
B4 0.153 none none
CI -0.153 +0.153 +0.153
AFTR none 0.153 0.153
[ H+1 ] = 0.153 M, [ OH-1 ] = 6.54E-14 M,
pH = 0.82 , pOH = 13.18
2. M(benzoic acid) = (3.582/122) / 0.30 = 0.0979 M
Shortcut[H+1] = sq.rt.(Ka *[HBz-x]) = 6.46E-6 M; OK.
[ OH-1 ] = 1.55E-9 M, pH = 5.19, pOH = 8.81
3. KOH, strong base, complete ionization (CI) (one way Rx):
KOH ---> K+1 + OH-1
B4 0.25 none none
CI -0.25 +0.25 +0.25
AFTR none 0.25 0.25
[ OH-1 ] = 0.25 M, [ H+1 ] = 4.0E-14 M, pOH = 0.60, pH = 13.40
4. a. A 3.0 % solution contains 3.0 g ammonia for every 97 g water.
mol ammonia = 3/17, volume solution = 97 mL, M = 1.82
b. Shortcut [OH-1]=sq.rt.(Kb *[NH3-x]) = 5.72E-3 M; OK.
[ H+1 ] = 1.75E-12 M, pH = 11.76, pOH = 2.24
5. the equilibrium is governed by the reaction:
HF (aq) = H+1(aq) + F-1(aq)
a. <--- b. ---> c. <--- d. --->
pH inc pH dec pH inc pH dec
6. a. conc. ammonium chloride = (10.0/53.5) / 0.50 = 0.374 M
is a salt, complete ionization so [ NH4+1 ] = 0.374 M
b. equilibrium is governed by weak base ammonia, NH3
NH3 + H2O = NH4+1 + OH-1
S 0.13 lots 0.374 none
C -x -x +x +x
E (0.13 - x) lots (0.374 + x) x
Shortcut [OH-1]=Kb * (0.13-x)/(0.374+x) = 6.26E-6 M; OK.
7. a. conc. sodium acetate = (7.65/82)/0.350 = 0.266 M
is a salt, complete ionization, so [ C2H3O2-1 ] = 0.266 M
b. equilibrium is governed by weak acetic acid, HC2H3O2
C2H3O2H = H+1 + C2H3O2-1
S 0.175 none 0.266
C -x +x +x
E (0.175 - x) x (0.266 + x)
Shortcut [H+1] = Ka * (0.175-x)/(0.266+x) = 1.18E-5 M; OK.
[ OH-1 ] = 8.44E-10 M, pH = 4.93, pOH = 9.07
8. a. concentrations after dilution:
(i) sodium hydroxide, (15*0.125)/90 = 0.0208 M
is a strong base, complete ionization, [ OH-1 ] = 0.0208 M
(ii) formic acid, (75*0.103)/90 = 0.0858 M
is a weak acid, it governs equilibrium
b. pre-equilibrium complete neutralization (CN) (one way Rx):
CHOOH + OH-1 ---> HCOO-1 + H2O
limiting rgt
B4 0.0858 0.0208 none lots
CN -0.0208 -0.0208 +0.0208 +0.0208
AFTR 0.0650 none 0.0208 lots
c. formic acid governs the equilibrium condition:
CHOOH = HCOO-1 + H+1
S 0.0650 0.0208 none
C -x +x x
E (0.0650 - x) (0.0208 + x) x
Shortcut[H+1]=Ka * (0.0650-x)/(0.0208+x)=6.56E-4 M; OK.
[ OH-1 ] = 1.52E-11 M, pH = 3.18, pOH = 10.82
9. a. concentrations after dilution:
(i) propionic acid, 1.00*1.50/2.00 = 0.075 M, a weak acid.
(ii) hydroiodic acid, 1.00*0.125/2.00 = 0.0625 M, a strong acid
b. perform some test calculations, find [ H+1 ] from each acid
as if it were the only acid present:
[ H+1 ] from hydroiodic acid = 0.0625 M
[ H+1 ] from propionic acid = sq.rt.(Ka*0.075) = 1.02E-3 M
c. compare the two [ H+1 ]:
0.0625 (from HI)
0.00102 (from propionic acid)
conclude that almost all solution acidity (98%) is due to HI,
and that from propionic acid (2%) may be neglected.
pH = 1.20
Send corrections, comments, and suggetions to S. F. Pavkovic