Question

[H+]= 5.5×10⁻³M .

Express the molarity to two significant digits.

[OH−] =

  M  

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Part B

[H+]= 1.5×10⁻⁹M .

Express the molarity to two significant digits.

[OH−] =

  M  

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Part C

A solution in which [H+] is 1000 times greater than [OH−].

Express the molarity to two significant digits.

[OH−] =

  M  

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Answer 1

Part A:

We know that Kw = [H⁺] [OH^-] = 1 x 10^-14

Given [H⁺]= 5.5x10^-3 M

[OH^-] = Kw / [H⁺] = 1x10^-14 / 5.5x10^-3 = 1.8 x10^-12 M.

Part B:

From the above equation, Kw = [H⁺] [OH^-] = 1 x 10^-14

Given [H⁺]= 1.5x10^-9 M

[OH^-] = Kw / [H⁺] = 1x10^-14 / 1.5x10^-9 = 6.6 x10^-6 M.

Part C:

Given [H⁺] is 1000 times greater than [OH^-] That means [H⁺]=1000[OH^-]

Kw = [H⁺] [OH^-] = 1 x 10^-14

[OH^-] = Kw / [H⁺] = 1x10^-14 /1000[OH^-]

[OH^-]²=0.1x10^-16

[OH^-] = (0.1x10^-16)^1/2

[OH^-] = 3.16x10^-9 M.