# How do you calculate the pH & pOH of a solution which has a hydroxide concentration of 0.0053 M?

Apr 8, 2016

$\text{pH} = 11.72$

$\text{pOH} = 2.28$

#### Explanation:

Your tool of choice here will be the relationship that exists between the pH and the pOH of an aqueous solution at room temperature

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Your strategy will be to use the definition of the pOH and the given concentration of hydroxide anions, ${\text{OH}}^{-}$, to find the pOH of the solution first, then use the above equation to get its pH.

So, the pOH of a solution is defined as

color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))

In your case, you know that the solution has

["OH"^(-)] = "0.0053 M"

Plug this value into the above equation to get

$\text{pOH} = - \log \left(0.0053\right) = \textcolor{g r e e n}{2.28}$

This means that you have

$\text{pH" = 14 - "pOH}$

$\text{pH} = 14 - 2.28 = \textcolor{g r e e n}{11.72}$

Notice that the result is consistent with the fact that the solution contains more hydroxide anions than hydronium cations, ${\text{H"_3"O}}^{+}$, i.e. you're dealing with a basic solution.

ALTERNATIVE APPROACH

You can also solve this problem by using the equation

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\left[{\text{H"_3"O"^(+)] * ["OH}}^{-}\right] = {10}^{- 14}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

to find the concentration of hydronium cations, ${\text{H"_3"O}}^{+}$. In this case, you will have - I won't add the units for the ionization constant of water, ${K}_{W} = {10}^{- 14}$

["H"_3"O"^(+)] = 10^(-14)/0.0053 = 1.887 * 10^(-12)"M"

Once again, the pH of the solution will be

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

$\text{pH} = - \log \left(1.887 \cdot {10}^{- 12}\right) = \textcolor{g r e e n}{11.72}$