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A) What is the change in pH when [OH-] increases by a factor of 10? b) What is the change in [H3O+] when the pH decreases by 3 units?

Chemistry

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2 parts
a

pH increases by 1 unit\text{pH increases by 1 unit}

b

increases by a factor of 1000\text{increases by a factor of 1000}

Step-by-step solution

7 steps

Here's how to determine the changes in pH and [H3O+][H_3O^+]:

Part a· 4 steps

What is the change in pH when [OH−][OH^-] increases by a factor of 10?

  1. Define the initial and final concentrations of [OH−][OH^-].

    Let the initial hydroxide ion concentration be [OH−]1[OH^-]_1.
    The final hydroxide ion concentration is [OH−]2=10×[OH−]1[OH^-]_2 = 10 \times [OH^-]_1.

  2. Relate pOH to [OH−][OH^-].

    The pOH is defined as pOH=−log⁡[OH−]pOH = -\log[OH^-].
    Initial pOH: pOH1=−log⁡[OH−]1pOH_1 = -\log[OH^-]_1
    Final pOH: pOH2=−log⁡(10×[OH−]1)=−(log⁡10+log⁡[OH−]1)=−1−log⁡[OH−]1=pOH1−1pOH_2 = -\log(10 \times [OH^-]_1) = -(\log 10 + \log[OH^-]_1) = -1 - \log[OH^-]_1 = pOH_1 - 1.
    So, the pOH decreases by 1 unit.

  3. Relate pH to pOH.

    At 25∘C25^\circ C, pH+pOH=14pH + pOH = 14.
    Initial pH: pH1=14−pOH1pH_1 = 14 - pOH_1
    Final pH: pH2=14−pOH2=14−(pOH1−1)=14−pOH1+1=pH1+1pH_2 = 14 - pOH_2 = 14 - (pOH_1 - 1) = 14 - pOH_1 + 1 = pH_1 + 1.

  4. Calculate the change in pH.

    The change in pH is ΔpH=pH2−pH1=(pH1+1)−pH1=1\Delta pH = pH_2 - pH_1 = (pH_1 + 1) - pH_1 = 1.
    The pH increases by 1 unit.

    The change in pH is: pH increases by 1 unit\boxed{\text{pH increases by 1 unit}}

Part b

increases by a factor of 1000\text{increases by a factor of 1000}

What is the change in [H3O+][H_3O^+] when the pH decreases by 3 units?

  1. Define the initial and final pH values.

    Let the initial pH be pH1pH_1.
    The final pH is pH2=pH1−3pH_2 = pH_1 - 3.

  2. Relate [H3O+][H_3O^+] to pH.

    The hydronium ion concentration is related to pH by the formula [H3O+]=10−pH[H_3O^+] = 10^{-pH}.
    Initial [H3O+][H_3O^+]: [H3O+]1=10−pH1[H_3O^+]_1 = 10^{-pH_1}
    Final [H3O+][H_3O^+]: [H3O+]2=10−pH2=10−(pH1−3)[H_3O^+]_2 = 10^{-pH_2} = 10^{-(pH_1 - 3)}

  3. Simplify the expression for the final [H3O+][H_3O^+].

    [H3O+]2=10−pH1+3=10−pH1×103[H_3O^+]_2 = 10^{-pH_1 + 3} = 10^{-pH_1} \times 10^3

    Substitute [H3O+]1=10−pH1[H_3O^+]_1 = 10^{-pH_1}:

    [H3O+]2=[H3O+]1×1000[H_3O^+]_2 = [H_3O^+]_1 \times 1000

    The change in [H3O+][H_3O^+] is: increases by a factor of 1000\boxed{\text{increases by a factor of 1000}}

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