# OH- Concentration of 0.015 M HCl?

Apr 6, 2018

$\approx 6.61 \times {10}^{-} 13 m o l {\mathrm{dm}}^{-} 3$

#### Explanation:

Assuming that this is done under standard conditions we can use the following relationship:

$p H + p O H = 14$ (at 25 degrees Celcius)

This is crucial since it connects $p H$ and $p O H$ of a given solution- lets say that the$p H$ is 1, then the $p O H$ will be 13.

Also remember that:

$p H = - \log \left[{H}_{3} {O}^{+}\right]$

$p O H = - \log \left[O {H}^{-}\right]$

In this problem, We can assume that $H C l$ is completely ionized in water because $H C l$ is a strong acid so the ${H}_{3} {O}^{+}$ concentration will also be $0.015 m o l {\mathrm{dm}}^{-} 3$

Then using the equation to find the pH of the given solution:

$p H = - \log \left[{H}_{3} {O}^{+}\right]$

$p H = - \log \left[0.015\right]$

$p H \approx 1.82$

$1.82 + p O H = 14$

$p O H \approx 12.18$

Now we must reverse-engineer the process.

If:

$p O H = - \log \left[O {H}^{-}\right]$

Then:

${10}^{- p O H} = \left[O {H}^{-}\right]$

Hence:

${10}^{-} \left(12.18\right) = \left[O {H}^{-}\right]$

$\left[O {H}^{-}\right] = {10}^{-} \left(12.18\right) \approx 6.61 \times {10}^{-} 13 m o l {\mathrm{dm}}^{-} 3$