# What would the concentration of hydroxide anions be in an aqueous solution that has ["H"_3"O"^(+)] = "0.325 M" ? Given: K_w = 10^(-14) "mol"^2"L"^2

Apr 21, 2016

$3.08 \times {10}^{- 14} \text{ ""mol""/""l}$

#### Explanation:

At ${25}^{\circ} \text{C}$ the ionic product of water is given by:

${K}_{w} = \left[{H}_{3} {O}_{\left(a q\right)}^{+}\right] \left[O {H}_{\left(a q\right)}^{-}\right] = {10}^{- 14} {\text{ ""mol"^(2)."l}}^{- 2}$

$\therefore \left[O {H}_{\left(a q\right)}^{-}\right] = {K}_{w} / \left[\left[{H}_{3} {O}_{\left(a q\right)}^{+}\right]\right]$

$\left[O {H}_{\left(a q\right)}^{-}\right] = {10}^{- 14} / 0.325 = 3.08 \times {10}^{- 14} \text{ ""mol""/""l}$