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1 August, 04:00

What volume of 0.850 M barium hydroxide solution do you need to make 3.00 L of a solution with a pH of 11.4002

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  1. 1 August, 04:10
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    Volume of 0.85M Ba (OH) ₂ needed = 4.45 ml and dilute to 3.00 Liters

    Explanation:

    ? Vol of 0.85M Ba (OH) ₂ = > 3.00 Liters solution with pH = 11.4002?

    pH = 11.4002 = > pOH = 14 - 11.4002 = 2.5998

    [OH⁻] = 10⁻²°⁵⁹⁹⁸ M = 2.513 x 10⁻³ M

    Since [OH⁻] is 2 times concentration of Ba (OH) ₂ then ...

    [Ba (OH) ₂] = 1/2[OH⁻] = 1/2 (2.513 x 10⁻³) M = 1.257 x 10⁻³ M

    Using Dilution Equation ...

    Molarity (concentrate) x Volume (concentrate) = Molarity (dilute) x Volume (dilute)

    => Vol (dilute needed) = M (c) x V (c) / M (d) = (1.251 x 10⁻³M) (3000ml) / (0.85M)

    = 4.45 ml needed

    Mixing = > 4.45 ml of 0.85M Ba (OH) ₂ + water solvent up to but not to exceed 3000ml.

    Test Calculation:

    Given 4.45ml of 0.85M Ba (OH) ₂ solution dilute to 3000ml and calculate pH.

    [Ba (OH) ₂] = moles Ba (OH) ₂/Vol Soln in Liters = 0.00445L x 0.85M / 3.00L = 0.0012608M = > [OH⁻] = 2 x 0.0012608M = 0.002523M in OH⁻

    pOH = - log[OH⁻] = - log (0.002523) = 2.598

    pH = 14 - 2.598 = 11.4017 (close enough)
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