What is the pH of a #1.4 * 10^-2 M# NaOH solution?
1 Answer
Explanation:
Even before doing any calculations, you can say that since you're dealing with a strong base, the pH of the solution must be higher than
The higher the concentration of the base, the higher the pH will be.
In your case, you're dealing with a solution of sodium hydroxide,
#"NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH"_text((aq])^(-)#
Notice that the salt dissociates in a
#["OH"^(-)] = ["NaOH"] = 1.4 * 10^(-2)"M"#
Now, the pH of the solution is determined by the concentration of hydronium ions,
#K_W = ["OH"^(-)] * ["H"_3"O"^(+)]#
At room temperature, you have
#K_W = 10^(-14)#
This means that the concentration of hydronium ions can be determined by using
#["H"_3"O"^(+)] = K_W/(["OH"^(-)])#
Plug in your values to get
#["H"_3"O"^(+)] = 10^(-14)/(1.4 * 10^(-2)) = 7.14 * 10^(-13)"M"#
The pH of the solution is equal to
#color(blue)("pH" = - log(["H"_3"O"^(+)])#
In your case,
#"pH" = -log(7.14 * 10^(-13)) = color(green)(12.15)#
As predicted, the pH is not only higher than
Alternatively, you can use the pOH of the solution to find its pH. As you know,
#color(blue)("pOH" = - log(["OH"^(-)]))#
In your case,
#"pOH" = - log(1.4 * 10^(-2)) = 1.85#
You know that
#color(blue)("pH" + "pOH" = 14)#
and so, once again, you have
#"pH" = 14 - 1.85 = color(green)(12.15)#