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    pH and [OH-] of Basic Solution

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    [OH-] and the pH for each of the following strong base solutions:

    -- 0.082 M KOH

    -- 1.065 g of KOH in 500.0 mL of solution

    -- 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL

    -- a solution formed by mixing 10.0 mL of 0.015 M Ba(OH)2 with 40.0 mL of 7.5x10-3 M NaOH.

    © BrainMass Inc. brainmass.com December 24, 2021, 7:44 pm ad1c9bdddf
    https://brainmass.com/chemistry/physical-chemistry/acid-base-equilibria-concentrations-ph-214935

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    Acid-Base Equilibria
    I am unsure of how to calculate

    [OH-] and the pH for each of the following strong base solutions:
    The pOH is calculated as pOH = - log[OH-]
    And pOH + pH = 14

    -- 0.082 M KOH
    In solution, KOH is a strong base and electrolyte, so
    KOH (aq) → K+ (aq) + OH- (aq)
    0.82 0.082
    Thus, the concentration of OH- is
    [OH-] = 0.082
    Then
    pOH = - log[OH-] = - log0.082 = 1.086
    pH = 14 - pOH = 14 - 1.086 = 12.914

    -- 1.065 g of KOH in 500.0 mL of solution
    the moles of KOH is 1.065 g / 56 g/mol = 0.019 mol
    So the concentration of 500.0 mL = 0.5 L solution is
    0.019 mol / 0.5 L = 0.038 M
    So the concentration of OH- is
    [OH-] = 0.038 M
    Then
    pOH = - log[OH-] = - log0.038 = 1.42
    pH = 14 - pOH = 14 - 1.42 = 12.58

    -- 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL
    Ca(OH)2 (aq) → Ca2+ (aq) + 2OH- (aq)
    Then [OH-] = 2 * 0.0105 = 0.021 M in the 10.0 mL solution.
    The mole of OH- is 10.0 * 0.021 = 0.21 mmol
    After the dilution, the concentration is then
    0.021 / 500.0 = 0.00042 M
    [OH-] = 0.00042 M
    pH = 14 + log[OH-] = 10.6

    -- a solution formed by mixing 10.0 mL of 0.015 M Ba(OH)2 with 40.0 mL of 7.5x10-3 M NaOH
    Ba(OH)2 (aq) → Ba2+ (aq) + 2OH- (aq)
    The concentration of OH- in this solution itself is 0.015 M * 2
    The moles of OH- from Ba(OH)2 is
    10.0 * 0.015 * 2 = 0.3 mmol

    The moles of OH- from NaOH is
    40.0 * 0.0075 = 0.3 mmol
    So the total OH- is 0.6 mmol
    The volume of the solution is 10.0 + 40.0 = 50.0 mL
    So
    [OH-] = 0.6 mmol / 50 mL = 0.012 M
    pH = 14 + log[OH-] = 12.08.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 7:44 pm ad1c9bdddf>
    https://brainmass.com/chemistry/physical-chemistry/acid-base-equilibria-concentrations-ph-214935

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