Warm-Up- Buffers and Titrations
Don’t Forget:
pH= -log[ H+ ]
pOH= -log [ OH- ]
H2O(l)  H+ (aq) + OH- (aq)
Kw= [H+][OH-] = 10-14 = KaKb
HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)
pH=pKa + log( [conjugate base]/[conjugate acid] )
pOH = pKb + log( [conjugate acid] / [weak base] )
For a weak acid buffer, when [conjugate base] = [weak acid] (in titration, halfway to equivalent
point) , the pH=_________pKa because the log1=0. Likewise, for a weak base buffer solution,
pOH=_______pKb when [conjugate acid]=[weak base]
Which of the following sets of materials would make the best buffer solution?
a.
b.
c.
d.
H2O, 1 M NaOH, 1 M H2SO4
H2O, 1 M HC2H3O2, 1 M NaC2H3O2
H2O, 1 M HC2H3O2, 6 M NaC2H3O2
H2O, 1 M HC2H3O2, 1 M NaOH
B is correct. A buffer solution is prepared from a weak acid and its conjugate
base, preferably in near-equal quantities. Choices A and D are wrong because
they do not show conjugate acid/base pairs. Choice C is wrong because it shows
a weak acid and its conjugate base, where the concentrations fo the acid and
the base are quite different. Thus, the best buffer solution would be that
prepared from Choice B, which shows a conjugate acid/base pair both present
in 1 M concentrations.
A certain buffer solution is 3 M in HF, and 2 M in NaF. Calculate the pH of this buffer given that
the Ka of HF=7.0x 10-4.
Ka= [H+][F-] / [HF]
7x10-4 = [H+](2) / 3
[H+] = 1.05 x 10-3
pH = -log [H+] = 2.98
Another way to solve this problem is by using the Henderson-Hasselbalch equation:
pH=pKa + log [conjugate base] / [weak acid]
 pH= -log Ka + log [F-] / [HF] = 3.155 – 0.176 = 2.98
Which of the following combination would produce a buffer solution of pH = 4? (K a HNO2 = 4.5
x 10-4)
a.
b.
c.
d.
0.3 M HNO2, 0.22 M NaNO2
0.22 M HNO2, 0.30 M NaNO2
0.11 M HNO2, 0.50 M NaNO2
050 M HNO2, 0.11 M NaNO2
The Henderson-Hasselbalch equation may again be used here:
pH= pKa + log [A-]/[HA]
4= 3.35 + log [A-] / [HA]
0.65 = log [A-]/[HA]
[A-]/[HA] = 4.5
Write out chemical equations describing the buffer activity which prevents drastic pH changes
when a.) strong acid and b.) strong base is added.
a.) Strong Acid:
a. Salt in buffer dissociates completely: NaX  Na+ + Xb. Added strong acid dissociates completely: HCl  H++ Clc. Protons from the acid are absorbed by the strong conjugate base of the
salt: H+ + X-  HX
b.) Strong Base:
a. Weak acid in buffer hardly dissociates: HX  HX
b. Added strong base dissociates completely: NaOH  Na+ + OHc. OH- is a strong base, which abstracts protons form the weak acid: OH- +
HX  H2O + XTitration is a procedure used to determine the molarity of an acid or base by reacting a known volume
of a solution of __________________unknown concentration with a known volume of a solution of known
concentration.
True/False: While a strong acid/strong base titration will have an equivalence point at pH 7, the
equivalence point need not always occur at pH 7. True, and when titrating polyprotic acids or
bases, there are several equivalence points, as each different acidic or basic species is titrated
separately.
Indicators are (strong or weak?) organic acids or bases that have different colors in their
undissociated and dissociated states. Indicators are used in (high or low?) concentrations and
therefore (do / do not?) significantly alter the equivalence point.
When a strong acid and a base or a strong base and an acid are placed in the same container, a
neutralization reaction occurs: HA + BOH  H2O + B+ + A-
This reaction will continue to produce water and salt until ______________ either the acid
(HA) or the base (BOH) is consumed
When there is an excess of acid in solution, the base will be entirely consumed in the
neutralization reaction, and the solution will be ___________acidic
-
When there is an excess _________of base in solution, the acid will be entirely consumed, and
the solution will be basic.
When the quantity of acid is equal to the quantity of base, the solution is completely
neutratlized and has a pH of __________ 7
What is the pH of a 100 mL aqueous solution containing 0.10 moles HCl and 0.07 moles NaOH?
a.)
b.)
c.)
d.)
0.52
4.52
8.52
12.52
A is correct. To solve this problem, we must recognize that we have an excess of 0.03
moles of the strong acid HCl. This translates to an excess of 0.03 moles H+. Therefore,
the hydrogen ion concentration is given by: [H+]= 0.03 mol / 0.100 L = 0.3 M.
Therefore, pH= -log [H+]= -log[0.3] = 0.52
As base is added to acid, the pH of the solution rises until all the acid has been consumed and
______________ excess base is present.
The equivalence point of a titration curve is the point at which exactly one equivalent of base has been
added for one equivalent of acid. At any point in a strong acid titration curve prior to the equivalence
point, the pH may be calculated from the concentration of __________acid. At any point after the
equivalence point, the pH is determined by the excess ______________base present.
100 mL of 0.2 M hydrochloric acid, a strong acid, is titrated with 0.2 M sodium hydroxide, a strong
base. What is the pH after a.) 50 mL, b.) 100 mL, and c.) 150 mL of NaOH have been added?
a.) 1.17
b.) Since 1 Hcl is neutralized for every 1 NaOH added, 0.02 moles of HCl have been
neutralized. We are at the equivalence point. The equivalence point of a strong acid
by a strong base is at pH = 7
c.) 12.60 Since 1 HCl is neutralized for every 1 NaOH added, all the HCl has been
neutralized and excess NaOH remains.
The pH at the equivalence point of the titration of a strong acid with a strong base is:
a.) less than 7
b.) equal to 7
c.) greater than 7
d.) possibly any one of the above answer choices
B is correct. By definition, pH at the equivalence point of the titration of a
strong acid with a strong base is equal to 7 because both the strong acid and
the strong base ionize completely in water. The equivalence point is the point
at which [H+]=[OH-], which is defined as pH=7
The titration of a weak acid by a strong base is vastly different from the titration of a strong acid by a
strong base: the equivalence point is not equal to pH=7, but is determined
by_______________________________ the pKa of the weak acid.
Consider the titration fo 100 mL of 0.100 mL of a 0.100 M solution of acetic acid (Ka=1.8 x 10-5) with a
0.100 M solution of sodium hydroxide. What is the pH a.)before the titration and after b.) 50 mL,
c.)100 mL, and d.) 150 mL of NaOH have been added?
a.)The pH of the solution prior to the titration must be calculated from the equilibrium
expression: CH3COOH  H+ + CH3COO- . Solve with Ka value and ice table the [H+], and
pH= -log[H+] = -log [1.34 x 10-3] = 2.87
b.) the pH of the solution after the addition of 50 mL of 0.100 M NaOH must be calculated
from the equilibrium expression: by the ice table. Therefore, after the addition, the
molarity of CH3COOH is : 0.005 mol / 0.150 L = 0.033 M, and the molarity of CH3COO- is
also 0.005 mol / .150 L = 0.033 M. Therefore, solving with Ka and concentrations,
[H+]=1.8x10-5. pH= 4.74
c.) The pH of the solution after the addition of 100 mL of 0.100 M NaOH is given by the ice
table, the molarity of CH3COO- is 0.010mol / .2 L = 0.05 M. In this case, must consider the
base ionization constant expression, Kb= [CH3COOH] [OH-] / [CH3COO-] = Kw / Ka=
1x10-14 / 1.8 x 10-5 = 5.6 x 10-10. Solve for OH- in ice table, [OH-]=5.29 x 10 -6 , pOH= log [OH-]= 5.28. pH= 8.72
d.) since NaOH is a much stronger base than the acetate ion, we can ignore OHcontribution made by the acetate ion once we are beyond the equivalnce point. The pH
after the addtion of 150 mL of base is calculated in a manner identical to that for the
titration of a strong acid ….pH= 12.30
The pH at the equivalence point of the titration of a strong acid with a weak base is:
a.) less than 7
b.) equal to 7
c.) greater than 7
d.) possibly any one of the above answer choices
A is correct. Since the acid, unlike the base, will dissociate completely in
solution, the pH, even in the presence of equivalent amount sof the acid and the
base, will favor the acid.
The two pKa values of sulfurous acid (H2SO3) are 1.81 and 6.91, respectively. The predominant species
at pH = 9 is:
a.) H2SO3
b.) HSO3c.) SO32d.) H+
C is correct. At a pH of 9, the hydrogen ion concentration is 1.x 10 -9. It is not likely
that a concentration of this magnitude is indicative of a predominate species, so
choice D is eliminated. At pH of 1.81, the titration of the first acidic proton is halfway
complete, and at a pH of 9, there is not likely to be a measureable quantity of H2SO3
left. Therefore, choice A is elminated. At pH of 6.91, the titration of the second acidic
proton is half way complete; at this point [HSO3-]=[SO32-] Therefore, for any pH
greater than 6.91, [SO3 2-] > [HSO3-], elinate choice B, which leaves only choice C.