Cliffs AP Chemistry, 3rd Edition

Step 3: Calculate the number of moles of OH–.

There should be twice as many moles of OH–as moles of Ba(OH) 2.

(OH)

.. (OH)
12 1.

0 100

1

0 100

mole Bamoles OH 0 0200

liter

liter

mole Ba

mole OH

2

2

## =

Step 4: Write the net ionic equation.

H++ OH–→H 2 O

Step 5: Because every mole of H+uses 1 mole of OH–, calculate the number of moles of excess
H+or OH–.

2.00 × 10 –2mole OH–−1.00 × 10 –2mole H+

= 1.00 × 10 –2mole OH–excess

Step 6: What is the approximate pH in the final solution?

pOH = –log[OH–] = −log[1.00 × 10 –2] = 2

pH = 14.00 −pOH = 14.00 −2.00 = 12.00

Another way to do this step, if you could use a calculator, would be

.

OH..

liter

mole M

0 600

• ==100 10# -^2 0 0167
  7A

pOH = 1.778

pH = 12.222

10. A student wants to make up 250 mL of an HNO 3 solution that has a pH of 2.00. How

many milliliters of the 2.00 M HNO 3 should the student use? (The remainder of the

solution is pure water.)

A. 0.50 mL

B. 0.75 mL

C. 1.0 mL

D. 1.3 mL

E. This can’t be done. The 2.00M acid is weaker than the solution required.

Answer: D

Step 1: Calculate the number of moles of H+in 250 mL of a HNO 3 solution which has a pH of
2.00. HNO 3 is a monoprotic acid. pH = −log[H+], so 2.00 = −log[H+] and [H+] = 1.00 × 10 –2M.

... mole H
1

100 10

1

025

mole H litermole H 25 10

#^2 liter 3

+==##

-+ -+

Acids and Bases