Question

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

Answer 1

`"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)`