Question

QUESTION 3

Calculate the pH for an aqueous solution that contains 2.15 x 10^-4 M hydroxide ion at equilibrium.

		4.65 × 10-11
		2.15 × 10-4
		3.67
		10.33

1 points (Extra Credit)   

QUESTION 4

What is the pH of a 0.020 M Ba(OH)2 solution?

		1.40
		1.70
		12.30
		12.60

Answer 1

3)

we have below equation to be used:

pOH = -log [OH-]

= -log (2.15*10^-4)

= 3.67

we have below equation to be used:

PH = 14 - pOH

= 14 - 3.67

= 10.33

Answer: 10.33

4)

[OH-] = 2*[Ba(OH)2] = 2*0.020 = 0.040 M

we have below equation to be used:

pOH = -log [OH-]

= -log (4*10^-2)

= 1.40

we have below equation to be used:

PH = 14 - pOH

= 14 - 1.40

= 12.60

Answer: 12.60