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Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

I am having trouble solving these problems. I'd appreciate a step by step explanation/tutorial if possible.?

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  • 2 months ago

    NaOH is a strong base so the concentration of OH- ions is the same as the concentration of the NaOH.

    The pOH of a solution is the negative logarithm of the OH- ion concentration:

    pOH = - log 0.0012 = 2.92

    The sum of pOH and pH is always 14 (at 25°C, technically). So:

    pH = 14 - pOH = 14 - 2.92 = 11.08

    The opposite of - log() is 10^-(), so when the pH is given the hydronium ion (H3O+) concentration is 10^-pH:

    10^-4.32 = 4.78 x 10^-5 M H3O+

    The product of the Hydronium ion concentration and the hydroxide ion concentration is always 10^-14 (at 25°C, technically).  So the hydroxide ion (OH-) concentration is:

    10^-14 / (4.78 x 10^-5) =  2.09 x 10^-10 M OH-

    In the case of Sr(OH)2 you need to be aware that both OH- ions dissociate, like this:

    Sr(OH)2 → Sr{2+} + 2 OH{-}

    Since two OH- ions form for every one molecule of Sr(OH)2 that dissociate, the concentration of OH- ions is twice the concentration of Sr(OH)2.

    Starting with what was given:

    pOH = 14 - 10.46 =  3.54

    10^-3.45 =  0.0003548 M OH-

    (0.0003548 M OH-) x (1 mol Sr(OH)2 / 2 mol OH-) = 0.000177 M Sr(OH)2 =

    0.000177 mol/L Sr(OH)2

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  • david
    Lv 7
    2 months ago

    0.0012M NaOH  ...  strong base that completely ionizes

      NaOH  -->  Na+  + OH-

      every NaOH forms the same number of OH-

    0.0012M NaOH forms  0.0012 OH-

      pOH = -log[OH-]  =  2.92

      pH = 14 - pOH = 11.08

    ----------------------------------------------------

    pH = 4.32

      hydronium = H3O+ ion

      pH = - log[H3O+]  <<<  algebra shows that 

      [H3O+] = 10^(-pH-)  =  10^-4.32  <<<  use a calculator

       [H3O+]  =  4.7863x10^-5 M

      pOH = 14 - pH = 9.68

       [OH-] = 10^(-pOH)  <<<  similar to pH

       [OH-] = 2.0893x10^-10 M

    ==================================

    Sr(OH)2  -->  Sr2+  +  2OH-

       pOH = 14 - pH = 3.54

      [OH-] = 10^-3.54  =  2.884x10^-4 M

     from the ball equation .. Sr(OH)2  is 1/2 the cons of OH-

       [Sr(OH)2]  =  1.442x10^-4 M  ...  you should know that M (polarity) is moles/Liter

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  • 2 months ago

    NaOH is a strong base releases OH- on a 1 to 1 basis

    so [OH-] conc of disolved base

    then

    pOH = - log[OH-]

    and 

    pH + pOH = 14.00

    pH = 14.00 - pOH

    pH = -log[H+]

    [H+] = 10^(-pH)

    then [OH-] = 1E-14 / [H+]

    use the above to find [OH-]

    then conc of Sr(OH)2 is half of that

    because each Sr(OH)2 has 2 OH-

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