Answers and Explanations
Equilibrium 235- B—An acid, any acid, will give a pH below 7;
thus, answers C–E are eliminated. A 0.1-molar
solution of a strong acid would have a pH of 1.
Acetic acid is not a strong acid, so answer A is
eliminated. - A—The Knearest 10−^8 will give a pH near 8. The
answer must involve the H 2 PO 4 −ion. - C—The generic Kais:
Ka=[H+][A−]/[HA] = 9 × 10 −^4 =x^2 /1.0 −x
x= 3 × 10 −^2 M and the percent dissociation is
(3 × 10 −^2 /1.0) ×100% =3%You will need to be able to do calculations at this
level without a calculator.- D—The two substances are not a conjugate
acid–base pair, so this is not a buffer. Both com-
pounds are salts of a strong base and a weak acid;
such salts are basic (pH >7). - C—The two substances constitute a conjugate
acid–base pair, so this is a buffer. The pH should
be near −log Ka1. This is about 2 (acid). - E—The two substances constitute a conjugate
acid–base pair, so this is a buffer. The pOH should
be near −log Kb. This is about 4. The pH would
be about 14 − 4 =10. - B—Any time an acid is added, the pH will drop.
The reaction of the weak base with the acid pro-
duces the conjugate acid of the weak base. The
combination of the weak base and its conjugate is
a buffer, so the pH will not change very much
until all the base is used. After all the base has
reacted, the pH will drop much more rapidly. The
equivalence point of a weak base–strong acid titra-
tion is always below 7 (only strong base–strong
acid titrations will give a pH of 7 at the equiva-
lence point). The value of pOH is equal to pKb
half-way to the equivalence point. - E—If pH =4.0, then [H+] = 1 × 10 −^4 =[A−], and
[HA] =0.30 − 1 × 10 −^4. The generic Kais [H+][A−]/
[HA], and when the values are entered into this
equation: (1 × 10 −^4 )^2 /0.30 =3.3 × 10 −^8. Since you
can estimate the answer, no actual calculations need
be done.
9. A—This is an acid-dissociation constant, thus
the solution must he acidic (pH <7). The pH of
a 0.010 M strong acid would be 2.0. This is not
a strong acid, so the pH must be above 2. - A—A is the salt of a strong acid and a weak base;
it is acidic. B and E are salts of a strong acid and
a strong base; they are neutral. C and D are salts
of a weak acid and a strong base; they are basic.
The lowest pH would be the acidic choice. - E—Sodium nitrite is a salt of a weak acid and a
strong base. Ions from strong bases, Na+in this
case, do not undergo hydrolysis, and do not affect
the pH. Ions from weak acids, NO 2 −in this case,
undergo hydrolysis to produce basic solutions. - B—The presence of a strong acid, HNO 3 ,
would make this the most acidic (lowest pH). - E—The weak acid and the weak base partially
cancel each other to give a nearly neutral solution. - C—Both A and C are buffers, because they have
conjugate acid–base pairs of either a weak acid
(A) or a weak base (C). The weak acid buffer
would have a pH below 7, and the weak base
buffer would have a pH above 7. - A—See the answer to question 14.
- A—The equilibrium constant expression is: Kb=
4.0 × 10 −^10 =[OH−][C 6 H 5 NH 3 +]/[C 6 H 5 NH 2 ].
This expression becomes: (x)(x)/(1.0 −x) =4.0 ×
10 −^10 , which simplifies to: x^2 /1.0 =4.0 × 10 −^10.
Taking the square root of each side gives: x=
2.0 × 10 −^5 =[OH−]. Since you can estimate the
answer, no actual calculations need be done. - E—The solubility-product constant expression
is: Ksp=[Zn^2 +][IO 3 −]^2 = 4 × 10 −^6. This may be
rearranged to: [IO 3 −]^2 = 4 × 10 −^6 /[Zn^2 +].
Inserting the desired zinc ion concentration
gives: [IO 3 −]^2 = 4 × 10 −^6 /(1 × 10 −^6 ) =4. Taking
the square root of each side leaves a desired IO 3 −
concentration of 2 M. 2 mol of KIO 3 must be
added to 1.00 L of solution to produce this con-
centration. Since you can estimate the answer,
no actual calculations need be done.