# What's the OH- concentration in 0.48M CH3CO2- (aq) (Kb of CH3CO2- is 5.6 x 10^-10)?

Apr 16, 2018

See below:

#### Explanation:

We set up the ${K}_{b}$ expression:

${K}_{b} = \frac{\left[O {H}^{-}\right] \times \left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]}$

And then we plug in our respective values:

$\left(5.6 \times {10}^{-} 10\right) = \frac{\left[O {H}^{-}\right] \times \left[C {H}_{3} C O O H\right]}{0.48}$

We can assume that the change in concentration of the base is negligible. Creating an ICE table will highlight this but the change in concentration of the base will be very small so we can assume that it is close to unchanged.

Also, as the $O {H}^{-}$ and $C {H}_{3} C O O H$ exists in a 1:1 ratio, we can say the following:

$\left(5.6 \times {10}^{-} 10\right) = {\left[O {H}^{-}\right]}^{2} / \left[0.48\right]$

${\left[O {H}^{-}\right]}^{2} = 2.688 \times {10}^{-} 10$

$\left[O {H}^{-}\right] \approx 1.64 \times {10}^{-} 5$