Question #1108e
1 Answer
Yes, the pH is equal to
Explanation:
This is a perfect example of a problem that wants to test your understanding of how the concentrations of hydronium and hydroxide ions determine a solution's pH.
When a problem gives a different value for water ion product,
As you know, the ion product of water is equal to
#color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)])#
At
#K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-12)#
SIDE NOTE I'll skip the units for the ion product constant, they are not important in this context
But you should also know that a neutral solution will always have equal concentrations of hydronium and hydroxide anions, which means that
#K_w = x * x = x^2 = 10^(-12)#
#x = ["H"_3"O"^(+)] = ["OH"^(-)] = sqrt(10^(-12)) = 10^(-6)"M"#
Now take a look at the molarity of that sodium hydroxide solution. The concentration of hydroxide anions in a
Why is that so?
Well, you know that at
In order to have a
But
In other words, you will have
#n_(OH^(-)) = overbrace(10^(-6)"moles OH"^(-))^(color(purple)("present in pure water")) + overbrace(10^(-9)"moles OH"^(-))^(color(purple)("added to the water"))#
#n_(OH^(-)) ~~ 10^(-6)"moles"#
In a
#["OH"^(-)] ~~ 10^(-6)"M"#
Of course, this translates to
#"pOH" = - log(10^(-6)) = 6#
and
#"pH" = 12 - 6 = color(green)(6)#
Remember, the solution is neutral at
#["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-6)"M"#
As a final note, chemistry is not about calculations. Focus on understanding what's going on first, then worry about integrals, logs, equations, formulas, derivatives, and things of that nature.
If you focus on calculations too much, you'll end up not seeing the forest for the trees.