Complete the following table with the needed [H3O +], [OH -], and pH, and classi
ID: 796998 • Letter: C
Question
Complete the following table with the needed [H3O +], [OH -], and pH, and classify the solution as acidic, basic or neutral.
[H3O +] [OH -] pH Classification
6.3 x 10^-4
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2.5 x 10^-9
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9.4 Basic
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6.8 x 10^-4
Explanation / Answer
1)
Solving for [OH^-]:
Kw = [H3O^+] [OH^-] ... rearrange the equation to solve for [OH^-]
[OH^-] = Kw / [H3O^+] = (1.0 x 10^-14) / (6.3 x 10^-4) = 1.59 x 10^-11 M
Solving for pH:
pH = -log[H3O^+] = -log(6.3 x 10^-4) = 3.20
Classification:
acidic because pH < 7
2)
Solving for [H3O^+]:
Kw = [H3O^+] [OH^-] ... rearrange the equation to solve for [H3O^+]
[H3O^+] = Kw / [OH^-] = (1.0 x 10^-14) / (2.5 x 10^-9) = 4.0 x 10^-6 M
Solving for pH:
pH = -log[H3O^+] = -log(4.0 x 10^-6) = 5.40
Classification:
acidic because pH < 7
3)
Solving for [H3O^+]:
pH = -log[H3O^+] ... rearrange the equation to solve for [H3O^+]
[H3O^+] = 10^-pH = 10^-9.4 = 3.98 x 10^-10 M
Solving for [OH^-]:
Kw = [OH^-] [H3O^+] ... rearrange the equation to solve for [OH^-]
[OH^-] = Kw / [H3O^+] = (1.0 x 10^-14) / (3.98 x 10^-10) = 2.5 x 10^-5 M
Classification:
basic because pH > 7
4)
Solving for [H3O^+]
Kw = [H3O^+] [OH^-] ... rearrange the equation to solve for [H3O^+]
[H3O^+] = Kw / [OH^-] = (1.0 x 10^-14) / (6.8 x 10^-4) = 1.47 x 10^-11 M
Solving for pH:
pH = -log[H3O^+] = -log(1.47 x 10^-11) = 10.8
Classification:
basic because pH > 7