# Question #09dee

Jan 14, 2015

If the Molarity concentration of an acid is $5.39 X {10}^{-} 12 M$ we can use this information to determine values of $\left[{H}_{3} {O}^{+}\right] , \left[O {H}^{-}\right] , \left[p H\right] \left(h \texttt{p} : / s o c r a t i c . \mathmr{and} \frac{g}{c} h e m i s t r \frac{y}{a} c i \mathrm{ds} - \mathmr{and} - b a s e \frac{s}{t} h e - p h - c o n c e p t\right) \mathmr{and} p O H$

The Molarity is the Concentration [ ]. While your question does not distinguish which concentration is represented I will assume for our purposes the the $5.39 x {10}^{-} 12 M$ is the $\left[{H}_{3} O +\right]$ concentration.

Therefore, $\left[{H}_{3} O +\right]$ = $5.39 x {10}^{-} 12 M$

To convert a Molarity concentration to pH we can calculate
$- \log \left[{H}_{3} O +\right] = p H$
$- \log \left[5.39 x {10}^{-} 12 M\right] = 11.27$

Therefore, $p H = 11.27$

Now, pH + pOH = 14, so pOH = 14 - pH
$14 - p H = p O H$
$14 - 11.27 = 2.73$

Therefore, $p O H = 2.73$

Then we can convert the $p O H \to \left[O {H}^{-}\right]$ by calculating
$a n t i \log \left(- p O H\right) = \left[O {H}^{-}\right]$
$a n t i \log \left(- 2.73\right) = 0.00186 M$
Moving the decimal gives us a value of
$1.86 x {10}^{-} 3 M$

Therefore $\left[O {H}^{-}\right] = 1.86 x {10}^{-} 3 M$

Some checkpoints to see if your calculations are correct.

The exponents of the concentrations should add up to -15.

-12 and -3 give us a value of -15

The pH should value of the pH and pOH should be less than the exponents of the concentrations.

pH = 11.27 < 12
pOH = 2.73 < 3