Can't seem to match the answer help

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SUMMARY

The concentration of Sr(OH)2 in a solution with a pH of 11 is 5.0 x 10^-4 M. To determine this, first calculate the pOH by subtracting the pH from 14, resulting in a pOH of 3. The concentration of hydroxide ions [OH-] is then found using the formula [OH-] = 10^-pOH, yielding 0.001 M. Since Sr(OH)2 dissociates into two hydroxide ions, the concentration of Sr(OH)2 is half that of [OH-], confirming the correct concentration as 0.0005 M.

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What is the concentration of Sr(OH)2 in a solution with a pH of 11?

The answer is SUPPOSED to be 5.0 x 10^-4...buuut i keep getting 0.001..
i didnt think this would be a complicated question.. all i did was found the pOH to be 3..and found the concentration of OH, giving me .001
what am i doing wrong?:S
 
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I'm not completely sure (I should be sure though! :P)

Equation is:

Sr(OH)2 + H2O <==> 2OH- + Sr((OH)O <---??

But anyways, the important thing is that there the hydroxide anion has a subscript of 2. You have to pay attention to the stoichiometric rations; therefore the molar ratio of the concentration of Sr(OH)2 to OH is 1:2.

Your 0.001 concentration of OH- present in solution is correct, but then you have to divide it to two to get the concentration of the Sr(OH)2.

0.001 \div 2 = 0.0005 or 5.0 \times 10^{-4}

-Art
 
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It's possible that you may be making a calculation error or using the wrong formula. The correct way to find the concentration of Sr(OH)2 in a solution with a pH of 11 is to first find the pOH, which is 14 - pH, giving you a pOH of 3. Then, use the formula [OH-] = 10^-pOH to find the concentration of hydroxide ions. In this case, [OH-] would be 10^-3 or 0.001. Since Sr(OH)2 dissociates into two hydroxide ions, the concentration of Sr(OH)2 would be half of the concentration of hydroxide ions, which is 0.0005 or 5.0 x 10^-4. Double check your calculations and make sure you are using the correct formula to get the correct answer.
 

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