Question #79774

Nov 29, 2016

$\left[H {O}^{-}\right] = 3.98 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$

Explanation:

$\left[{H}_{3} {O}^{+}\right] = \frac{3.00 \times {10}^{-} 3 \cdot m o l}{12 \cdot L} = 2.5 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = - {\log}_{10} \left(2.5 \times {10}^{-} 4\right) = - \left(- 3.60\right) = 3.60$

Since $p H + p O H = 14$, $p O H = 10.40$,

And now we take anti-logs.......

$\left[H {O}^{-}\right] = {10}^{- 10.4} \cdot m o l \cdot {L}^{-} 1 = 3.98 \times {10}^{-} 11 \cdot m o l \cdot {L}^{-} 1$

Hydroxide ion concentration is low, because this is an acidic solution.