How do you calculate the pH & pOH of a solution which has a hydroxide concentration of 0.0053 M?
1 Answer
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,
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,
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"^(+)] = 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)#