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    pH at the equilibrium point

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    6) When 12.0ml of 0.250 M NaOH is added to 10.0ml of 0310 M HCL, has equilibrium been reached? Explain.
    What is the pH at this point ?

    © BrainMass Inc. brainmass.com December 24, 2021, 7:14 pm ad1c9bdddf
    https://brainmass.com/chemistry/acids-and-bases/ph-equilibrium-point-173733

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    acid base 6
    ..........................................................................................................
    6) Strong acid - strong base
    When 12.0ml of 0.250 M NaOH is added to 10.0ml of 0310 M HCL , has equilibrium
    been reached ? EXPLAIN
    What is the pH at this point ?

    First write down the chemical reaction equation:
    For the strong acid and strong base, they completely ionized in the solution.
    NaOH  Na+ + OH-
    HCl  Cl- + H+
    So
    NaOH + HCl  NaCl + H2O

    12.0 ml of 0.250 M NaOH is 12.0 * 0.250 = 3 mmol of NaOH
    10.0 ml of 0.310 M HCl is 10.0 * 0.310 = 3.1 mmol of HCl.
    From the above reaction, 1 mol of NaOH reacts with 1 mol of HCl to reach the equilibrium since NaCl is a neutral salt.

    Therefore, 3 mmol of NaOH reacts with 3 mmol of HCl. And 3.1 - 3 = 0.1 mmol HCl is left.
    NaOH + HCl  NaCl + H2O
    Beginning 3 mmol 3.1 mmol
    Equilibrium 0 0.1 mmol 3 mmol

    So the equilibrium has not been reached.
    There is 0.1 mmol HCl in the solution.
    The total volume is 12.0 + 10.0 = 22.0 ml.
    So the concentration of H+ is
    x = 0.1 mmol / 22.0 ml = 0.004545 M.
    the pH of the solution is
    pH = -log[H+] = -logx = -log 0.004545 = 2.34.

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

    © BrainMass Inc. brainmass.com December 24, 2021, 7:14 pm ad1c9bdddf>
    https://brainmass.com/chemistry/acids-and-bases/ph-equilibrium-point-173733

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