Question

Question #09dee

Answer 1

If the Molarity concentration of an acid is `5.39 X 10^-12 M` we can use this information to determine values of `[H_3O^+], [OH^-], [pH](http://socratic.org/chemistry/acids-and-bases/the-ph-concept) and pOH`

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 `[H_3O+]` concentration.

Therefore, `[H_3O+]` = `5.39 x 10^-12 M`

To convert a Molarity concentration to pH we can calculate
`-log[H_3O+] = pH`
`-log [5.39 x 10^-12 M] = 11.27`

Therefore, `pH = 11.27`

Now, pH + pOH = 14, so pOH = 14 - pH
`14 - pH = pOH`
`14 - 11.27 = 2.73`

Therefore, `pOH = 2.73`

Then we can convert the `pOH to [OH^-]` by calculating
`antilog (-pOH) = [OH^-]`
`antilog(-2.73) = 0.00186 M`
Moving the decimal gives us a value of
`1.86 x 10^-3 M`

Therefore `[OH^-] = 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

SMARTERTEACHER YOUTUBE