# What is the solution to the question; The hydrogen ion concentration of a sample of orange juice is 2.0×10^-11mol dm^-3. What is its "pOH"?

Mar 9, 2018

$\text{pOH} = 3.30$
$\text{pOH} = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left({K}_{w} / \left(\left[{H}^{+}\right]\right)\right)$
${K}_{w} = 1 \cdot {10}^{-} 14$
$\left[{H}^{+}\right] = 2.0 \cdot {10}^{-} 11$
$\text{pOH} = - \log \left(\frac{1 \cdot {10}^{-} 14}{2.0 \cdot {10}^{-} 11}\right) = 3.30$