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6 May, 12:36

A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added to the base?

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  1. 6 May, 12:44
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    Answer;

    pH = 12.33

    Explanation;

    The equation of reaction is:

    LiOH (aq) + HCl (aq) - -> LiCl (aq) + H2O (l)

    Reactants left after the titrant is added;

    Total Moles LiOH;

    = 0.035L LiOH * (0.2moles/L)

    = 0.007moles of LiOH

    Moles of HCl;

    = 0.023L HCl * (0.25moles/L)

    = 0.00575moles HCl is the limiting reagent

    Reacting amount of moles of LiOH;

    = 0.0575 moles HCl * (1mole LiOH/1moles HCl)

    =0.00575 moles LiOH (reacted)

    Moles of LiOH left;

    = 0.007moles total - 0.00575moles that react

    =.00125 moles of LiOH (left)

    LiOH is a strong base, which means that it ionizes completely.

    0.00125moles LiOH * (moles/0.058L) = 0.02155M of LiOH

    LiOH (aq) - -> Li + (aq) + OH - (aq)

    [LiOH] = [OH-] = 0.02155 M

    pOH = - log[OH-]

    pOH = - log (0.02155)

    pOH = 1.67

    pH = 14 - pOH

    pH = 14 - 1.67

    pH = 12.33
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