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60. The correct answer is (B). A Brönsted base is a proton acceptor. H
2
O is the proton acceptor in the
forward reaction, and PO
4
–3
is the proton acceptor in the reverse reaction.

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Lesson 10
302
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ARCO
■ SAT II Subject Tests
61. The correct answer is (A). Since this problem involves a strong acid and strong base, one can
assume that they dissociate completely. The normality of H
2
SO
4
is 2 and that of KOH is 1.
.01l 
× lM [H
+
] 
× 2 – 0.1l × .1M [OH
–
] = .01moles of H
+
.01moles of H
+
/.11l = .091M H
+
pH = 1.04
62. The correct answer is (E). If four moles of CO
2
are consumed, then 160kcal of heat are produced.
63. The correct answer is (D). Using the Henderson-Hasselbach equation:
pH = pKa + log [conj. base]/[acid]
6 = 5 + log [conj. base]/[acid]
log[base]/[acid] = 1
[base]/[acid] = 10
64. The correct answer is (A). This problem deals with the combustion of ethane; the products of such
a reaction are CO
2
, H
2
O, and heat.
65. The correct answer is (D). Utilizing the definition of an equilibrium constant, one can write:
K
eq
= [C] · [D]
3
/[A]
2
· [B]
66. The correct answer is (C). Increasing the temperature of an exothermic reaction favors the reverse
reaction. Raising the concentration of X increases the number of collisions per unit time thereby
increasing the rate of the forward reaction. A catalyst lowers the activation energy of the reaction, so
it increases the rate of the forward reaction.
67. The correct answer is (C). Atomic number equals the number of protons. Atomic mass equals the
number of protons plus neutrons. Atomic number less the charge equals the number of electrons.