Chemistry Reference
In-Depth Information
5. [n. conjugate base]—NH
2
-
is the conjugate base of NH
3
.
6. [A. acidic]—Remember: Some texts will concentrate on H
+
ions.
7. [B. HNO
3
]—Arrhenius acids must have H
+
ions to release.
8. [D. Ba(OH)
2
]—Arrhenius bases must have OH
-
ions to release.
9. [A. NH
3
]—Ammonia can't release OH
-
ions the way an Arrhenius base can,
but it can accept protons as a Brønsted-Lowry base.
10. [A. more H
3
O
+
than OH
-
]—Having more H
3
O
+
than OH
-
is a characteristic
of any acidic solution.
11. [C. remains the same]—k
w
is a constant for a given temperature.
12. [D. 1.0 × 10
-5
M]—Remember: -log(1.0 × 10
-5
) = 5.
13. [C. 1.0 × 10
-12
M]—If the pH is 2, then the concentration of hydronium
(H
3
O
+
) ions is 1.2 × 10
-2
M. We can determine the concentration of OH
-
ions by remembering that [H
3
O
+
] × [OH
-
] = 1.0 × 10
-14
.
k
w
[H
3
O
+
]
1.0 × 10
-14
1.0 × 10
-2
[OH
-
] =
=
= 1.0 × 10
-12
14. [A. 3.2]—
pH = -log[H
3
O
+
] = -log(5.64 × 10
-4
) = 3.248720896, which rounds to 3.2
15. [B. 3.4]—This question gives us the concentration of hydroxide (OH
-
) ions,
so we must solve as shown here:
pH = 14 - (-log[OH
-
]) = 14 - (-log(2.8 × 10
-11
)) = 3.4, after rounding
16. [C. 10]—Remember that pH + pOH = 14. We can solve for pH as follows:
pH = -log[H
3
O
+
] = -log(1.0 × 10
-4
) = 4. Because the question asks for the
pOH value, we solve pOH = 14 - pH = 14 - 4 = 10.
17. [B. basic]—Phenolphthalein is an indicator that turns pink in basic
solutions. It is often used to signal the endpoint of a titration.
18. [A. K
b
= 1.0 × 10
-3
]—Simply look for the greatest value for K
b
.
19. [D. K
b
= 1.0 × 10
-9
]—The weakest base has the strongest conjugate acid.
20. [B. NH
3
]—NH
4
+
NH
3
+ H
+
. The conjugate base is the species that is left
after the acid gives up one H
+
.
21. [4.0 L]—The H
2
SO
4
will release 2 moles of H
+
per mole of acid, so we will
multiply the molarity of the acid by 2 and solve as shown here:
Given: M
a
= 6.0 M; V
a
= 2.0 L; M
b
= 3.0 M
Find: V
b
V
a
M
a
M
b
(2.0 L)(6.0 M)
(3.0 M)
V
b
=
=
= 4.0 L
22. [130 ml]—Work shown following:
Given: M
a
= 1.0 M; V
a
= 250 ml; M
b
= 2.0 M
Find: V
b
V
a
M
a
M
b
(250 ml)(1.0 M)
(2.0 M)
V
b
=
=
= 125 ml = 130 ml, after rounding
