2) What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 x 10-14? 2) C) 7.000 A) 8.446 B) 6.767 D) 7.233 E) 0.465 3) A solution with a hydrogen ion concentration of 3.25 x 10-6 M is and has a hydroxide ion concentration of A) basic, 3.08x 10-8 M B) acidic, 3.08 D) basic, 3.08 10-8 M C) acidic, 3.08 10-9 M 10-9 M 4) A solution with a hydroxide ion concentration of 4.15 10-6 M is and has a hydrogen ion concentration of A) basic, 2.41 x 10-8 M C) acidic, 2.41 x 10-9 M B) basic, 2.41x 10-9 M D) acidic, 2.41 x 10-8 M

Formulas to be used:

[H+] *[OH-] = Kw

pH = – log [H+]

And

[H+] = [OH-] ( note : this formula is

valid only for pure water)

In general Kw = 10-14 (unless specified this is

always used)

2)B) 6.767

Explaination:

Given : pure water and Kw = 2.92 x 10-14

[H+] *[OH-] = Kw

(Pure water ,so [H+] = [OH-] )

[H+] *[H+] = Kw

[H+]2 = Kw

[H+]2 = 2.92 x 10-14

[H+] = (2.92 x 10-14 )1/2

[H+] =1.7088 x 10-7

pH = -log[H+]

pH = -log(1.7088 x 10-7)

pH = 6.767

3)C)acidic ,3.08 x 10-9M

Explaination:

Given : [H+] =3.25 x 10-6M

For calculation of pH :

pH = -log[H+]

pH = -log (3.25 x 10-6)

pH= 5.48

pH<7 , therefore , acidic
For [OH-] calculation:
[H+] *[OH-] = Kw
(3.25 x 10-6)*[OH-] = 10-14
[OH-] = (10-14 )/(3.25 x
10-6)
[OH-] = 3.08 x
10-9M
4) B) basic , 2.41 x 10-9
Explaination:
Given : [OH-] = 4.15 x 10-6M
For calculation of [H+] :
[H+]
*[OH-] = Kw
[H+] *(4.15 x 10-6) = 10-14
[H+] = (10-14 )/(4.15 x
10-6)
[H+] = 0.2409 x 10-8
[H+] = 2.41 x 10-9M
For pH calculation :
pH = -log[H+]
pH = - log(2.41 x 10-9)
pH = 8.61
pH>7 , therefore , basic