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    Acid-Base Equilibria Calculated

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    1. Calculate [OH-] for 2.170 g of LiOH in 300.0 mL of solution. [OH-] = ? M

    2. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.160 L to form a solution with a pH of 3.30?

    3. Calculate the molar concentration of [OH-]ions in a 7.0×10−2 M solution of ethylamine (C2H5NH2) (Kb= 6.4 x 10-4).

    4. Calculate [OH-] and pH for a solution of 7.0 x 10-2 M NaCN? I am not sure if I need Kb or Ka for this equation.

    5. What is the pH of a solution that is 3.0×10−9 M in NaOH? Would I do the log (3.0 x 10-9) and then subtract that number from 14 in order to get the pH?

    6. How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 10.0 L of a solution that has a pH of 1.99?

    Your help is greatly appreciated!

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    https://brainmass.com/chemistry/acids-and-bases/acid-base-equilibria-calculated-274534

    Solution Preview

    1. Calculate [OH-] for 2.170 g of LiOH in 300.0 mL of solution. [OH-] = ? M
    Compute for the Molarity of LiOH:
    M = moles LiOH / liter solution
    M = (mass/molar mass) / liter of solution
    M = (2.170 g/23.94 g) / 0.3 L
    M = 0.302
    Since LiOH is a strong base, it will dissociate 100% in solution. Hence:
    LiOH → Li(aq) + OH-(aq)
    0.302 M 0.302 M 0.302 M
    Therefore, [OH-] = 0.302 M

    2. How many moles of HF (Ka = 6.8 x 10-4) must be present in 0.160 L to form a solution with a pH of 3.30?

    HF → H+(aq) + F-(aq)
    The ratio of H+ and F- per mole of HF that dissociates is 1:1. Hence,
    [H+] = [F-]
    [H+] = antilog (-pH)
    [H+] = antilog (-3.30)
    [H+] = 5.01 x 10-4
    [F-] = 5.01 x 10-4
    Ka = [H+] [F-] / [HF] - [H+]
    6.8 x 10-4 = (5.01 x 10-4)( 5.01 x 10-4) / [HF] - 5.01 x 10-4
    [HF] = 8.71 x 10-4 M
    M = moles/liter solution
    8.71 x 10-4 M = moles HF / 0.16 L
    Moles HF = ...

    Solution Summary

    The expert calculates acid-base equilibrias. The pH of the solution are determined.

    $2.49

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