Question

# Calculate [H3O+][H3O+] and [OH−][OH−] for each of the following solutions at 25 ∘C∘C given the pHpH\rm...

Calculate [H3O+][H3O+] and [OH−][OH−] for each of the following solutions at 25 ∘C∘C given the pHpH\rm pH.

You may want to reference (Pages 696 - 700) Section 16.6 while completing this problem.

Part A

pH=pH= 8.53

[H3O+], [OH−] =

M

Part B

pH=pH= 11.32

[H3O+], [OH−] =

M

Part C

pH=pH= 2.94

[H3O+], [OH−] =

M

Some useful formula

pH + pOH = 14

pH = - log [H3O+]

So, [H3O+] = 10^(-pH)

pOH = - log [OH-]

So, [OH-] = 10^(-pOH)

Concentration of [H3O+] and [OH-] can be determined just using above formula direct

Part A :

pH = 8.53

[H3O+] = 10^(-pH ) = 10^(-8.53) = 2.95 * 10^(-9) M

pOH = 14 - pH = 14 - 8.53 = 5.47

[OH-] = 10^(-pOH) = 10^(-5.47) = 3.39 * 10^(-6) M

[H3O+] , [OH-] = 2.95 * 10^(-9) , 3.39 * 10^(-6) M

Part B :

pH = 11.32

[H3O+] = 10^(-pH) = 10^(-11.32) = 4.79 * 10^(-12) M

pOH = 14 - pH = 14 - 11.32 = 2.68

[OH-] = 10^(-pOH) = 10^(-2.68) = 2.09 * 10^(-3) M

[H3O+] , [OH-] = 4.79 * 10^(-12) , 2.09 * 10^(-3) M

Part C :

pH = 2.94

[H3O+] = 10^(-pH) = 10^(-2.94) = 1.15 * 10^(-3) M

pOH = 14 - pH = 14 - 2.94 = 11.06

[OH-] = 10^(-pOH ) = 10^(-11.06) = 8.71 * 10^(-12) M

[H3O+] , [OH-] = 1.15 * 10^(-3) , 8.71 * 10^(-12) M

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