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

1 Answer
Apr 8, 2016

Answer:

#"pH" = 11.72#

#"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

#color(purple)(|bar(ul(color(white)(a/a)color(black)("pH " + " pOH" = 14)color(white)(a/a)|)))#

Your strategy will be to use the definition of the pOH and the given concentration of hydroxide anions, #"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

#"pOH" = - log(0.0053) = color(green)(2.28)#

This means that you have

#"pH" = 14 - "pOH"#

#"pH" = 14 - 2.28 = color(green)(11.72)#

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

ALTERNATIVE APPROACH

You can also solve this problem by using the equation

#color(purple)(|bar(ul(color(white)(a/a)color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14))color(white)(a/a)|)))#

to find the concentration of hydronium cations, #"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)|)))#

#"pH" = - log(1.887 * 10^(-12)) = color(green)(11.72)#