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**Problem**: Calculate pH for 1.6x10-3 M Sr(OH)2.

###### FREE Expert Solution

###### FREE Expert Solution

Sr(OH)_{2 }→ Sr^{2+} + 2 OH^{-}

$\left[{\mathrm{OH}}^{-}\right]\mathbf{}\mathbf{=}\mathbf{}(1.6x{10}^{-3}\frac{\overline{)\mathrm{mol}\mathrm{Sr}{\left(\mathrm{OH}\right)}_{2}}}{1L})\mathbf{}\left(\frac{2\mathrm{mol}{\mathrm{OH}}^{-}}{1\overline{)\mathrm{mol}\mathrm{Sr}{\left(\mathrm{OH}\right)}_{2}}}\right)$

[OH-] = 3.2x10^{-3 }M

$\overline{){\mathbf{pOH}}{\mathbf{\hspace{0.17em}}}{\mathbf{=}}{\mathbf{}}{\mathbf{-}}{\mathbf{log}}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}$

pOH = -log (3.2x10^{-3 }M)

###### Problem Details

Calculate pH for 1.6x10^{-3} M Sr(OH)_{2}.

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Strong Acid-Base Calculations concept. If you need more Strong Acid-Base Calculations practice, you can also practice Strong Acid-Base Calculations practice problems .

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Based on our data, we think this problem is relevant for Professor Ratliff's class at USF.