Question #1108e

1 Answer
Stefan V.
Feb 20, 2016

Yes, the pH is equal to 66.

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, K_WKW, your first impulse should be to figure out the concentrations of hydroxide and hydronium ions present in a neutral sample of pure water at that given temperature.

As you know, the ion product of water is equal to

color(blue)(K_W = ["H"_3"O"^(+)] * ["OH"^(-)])KW=[H3O+]⋅[OH−]

At 80^@"C"80∘C, you know that this will be equal to

K_W = ["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-12)KW=[H3O+]⋅[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)Kw=x⋅x=x2=10−12

x = ["H"_3"O"^(+)] = ["OH"^(-)] = sqrt(10^(-12)) = 10^(-6)"M"x=[H3O+]=[OH−]=√10−12=10−6M

Now take a look at the molarity of that sodium hydroxide solution. The concentration of hydroxide anions in a "1 dm"^31 dm3 sample of that is so small that you can actually consider it negligible.

Why is that so?

Well, you know that at 80^@"C"80∘C, "1 dm"^31 dm3 of pure water contains 10^(-6)10−6 moles of hydroxide anions. Remember, when you're working with a "1-dm"^31-dm3 sample, molarity and number of moles are interchangeable.

In order to have a 10^(-9)"M"10−9M solution of sodium hydroxide, you need to add 10^(-9)10−9 moles of sodium hydroxide in 1 dm"^31dm3 of pure water.

But 10^(-9)10−9 moles of hydroxide anions is a very small amount compared with 10^(-6)10−6 moles of hydroxide anions that are already present in the pure water.

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 "1-dm"^3 sample, this will give you

["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 "pH" = 6 at 80^@"C", since you have

["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.

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