Using the fact that pKa(HF) = 3.17; show THREE ways to construct a BUFFER solution at pH=3.17
using the following:
0.50M solution of NaOH
0.25M solution of NaF
0.50M solution of HCl
0.25M solution of NaCl
0.50M solution of HF
1) One Volume of HF and 2 Volumes of NaF
2) One Volume of HF and ½ that volume of NaOH making F3) Two Volume of NaF and ½ volume of HCl to make HF
What is the pH of an aqueous solution of 0.754 M pyridine (a weak base with the formula
C5H5N) pKb= 8.79
9.48
8.92
4.47
9.54
none
BASE: pOH= ½(Kb-log(0.754) = 4.47 
pH= 9.54
What concentration of Calcium hydroxide (assume STRONG and it all dissolves) is needed to give an
aqueous solution with a pH of 10.730
2.69x10-4
5.37x10-4
1.34x10-4
3.27x10-4
none
pOH=
14-10.73
[OH] =
=10^-3.27
3.27
0.000537032
but [Ca(OH)2 ] = 0.5[OH]=
2.69E-04
If the pH of a solution of NaF is adjusted for pH. Using the fact that pKa(HF) = 3.17; at what
pH is 85% of the total fluoride in the form of F-?
3.17
3.92
2.42
=3.17LOG(0.15/0.85)
3.99
none
3.92
Which of the following aqueous solutions are good buffer systems ?
0.18 M hydrocyanic acid + 0.19 M sodium acetate
0.24 M hydrobromic acid + 0.22 M potassium bromide
0.04 M sodium fluoride + 0.47 M hydrofluoric acid
0.31 M ammonia + 0.34 M amonium chloride
0.36 M potassium bromide + 0.26 M barium bromide
NO different species
NO(strong acid)
NO(could be BUT KF too small
YES
NO(both salts)
What is the pH of a solution contains 0.482 M ammonium iodide and 0.589 M ammonia.
(pKb(NH3)= 4.75)
4.66 4.83 9.34 9.56 none
pH of BASE BUFFER:
pH= 9.34
pOH= pKb - log ([B]/[BH+]) = 4.75 - log(0.589/0.482) = 4.66 
A buffer solution contains 0.481 M nitrous acid and 0.314 M sodium nitrite. Ka = 4.50 x 10-4
If 0.0309 moles of sodium hydroxide are added to 250 ml of this buffer, what is the pH of the
resulting solution?
3.91
1.56
3.16
3.43
none
4.50E-04
3.347
0.12025
0.0785
0.0309
0.08935
0.1094
What is the Hydroxide ion concentration of an aqueous solution of 0.541 M tri
weak base with the formula (HOC3H6)3N)? Kb= 4.80x10-8
2.50x10-8
[OH- ]=
1.61x10-4
3.85x10-7
=
SQRT(0.541*0.000000048)
6.21x10-11
3.43
propanolamine (a
none
1.61E-04
What is the pH of a solution contains 0.202 M methylamine iodide and 0.389 M methylamine.
(pKb(CH3NH2)= 3.37)
10.35 3.65 3.09 10.91 10.60 none
pH of BASE BUFFER: pOH= pKb - log ([B]/[BH+]) = 3.37 - log(0.389/0.202) = 3.09 
pH= 10.91
200ml 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?
2.86
11.14 3.32
3.11
3.84
HA
A
SB
none
0.0962
0.0314
0.03
0.0662
0.0614
3.32
100ml of 0.25M Potassium Hydroxide is titrated to equivalence with 0.35M Sulfuric Acid, what
volume is required and what will be the final pH?
0.1*0.25=0.025/(2*0.35) => 35.7ml at pH=7
0.20 M
0.30 M
ACID
BASE
Equation/work
1.90
50 ml
Chloroacetic
Acid
(pKa=2.85)
15 ml
Ca(OH)2
50 ml NH3
(pKb=4.75)
3.34
2.76
8.95
9.55 9.43
moles of acid = MaVa
moles of base = MbVb
pOH=4.75LOG(0.005/0.01)=
5.05
2.80
0.010
0.009
moles of acid = MaVa
moles of base = MbVb
Weak Acid (HH Eq)
pH= 2.85log((0.001/0.009)
5.05
50 ml HCl
3.80
3.80
9.07
8.25
0.010
0.015
5.051029996
pH=
8.95