# Question 36e0b

Feb 15, 2017

$\text{pH} = 11$

#### Explanation:

The thing to remember about an aqueous solution at room temperature is that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\left[{\text{OH"^(-)] * ["H"_ 3"O}}^{+}\right] = {10}^{- 14}}}}$

Rearrange this equation to isolate the concentration of hydronium cations on one side

["H"_ 3"O"^(+)] = (10^(-14))/(["OH"^(-)])

Now, you should be aware that

color(blue)(ul(color(black)("pH" = - log( ["H"_3"O"^(+)]))))

Plug the expression you have for the concentration of the hydronium cations in this equation to get

"pH" = - log( (10^(-14))/(["OH"^(-)]))#

In your case, you will have

$\text{pH} = - \log \left(\frac{{10}^{- 14}}{{10}^{- 3}}\right)$

$\text{pH} = - \log \left({10}^{- 11}\right)$

$\text{pH} = - \left(- 11\right) \log \left(10\right)$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{pH} = 11}}}$