[chanelo]

The question states: If the Ka of a monoprotic weak acid is 8.6 x 10^-6, what is the pH of a 0.42 M solution of the acid?

What I did was:
(8.6 x 10^-6)(0.42) = [H⁺]
[H⁺] = 0.000003612
pH = -log[0.000003612]
pH = 5.44

But this is wrong, so can someone help me figure out where I messed up?

[Borek]

(8.6 x 10^-6)(0.42) = [H⁺]

No idea what you are doing here, it has nothing to do with the way such problems are solved.

Perhaps you forgot the square root? If so, you should also check if the condition that allows using this approximate equation holds.

[SomeDude]

imagine the theoretical monoprotic acid HA where H is hydrogen and A is a variable.

HA<=H₂0=>H⁺ + A^-

Ka =  ([H⁺][A])/[HA]

use the ICE (initial change equilibrium) technique and probably the quadratic formula if the acid has enough strength to affect [HA] when dissociated, to find [H], [A], and [HA] in the solution when [HA]₀ is 0.42M.

What i think you did was to skip the ICE step, you have to remember that a .42M solution of an acid means it is diluted, and that the H⁺ ions which dissociate will determine the pH.

have fun typing quadratic formula solutions into your calculator!

[Borek]

you have to remember that a .42M solution of an acid means it is diluted

It doesn't mean anything like that. "Diluted" is a relative term without a solid definition. In some cases 0.42M will be diluted, in some cases it will be quite concentrated, depends on the acid and circumstances.

[Big-Daddy]

(8.6 x 10^-6)(0.42) = [H⁺]

No idea what you are doing here, it has nothing to do with the way such problems are solved.

Perhaps you forgot the square root? If so, you should also check if the condition that allows using this approximate equation holds.

How would you intuitively check that an approximation holds? It's always better to use the exact calculation obviously, but some of us are training for exams and so approximations would save a great deal of time.

In this example, I would judge Ca as being easily high enough for virtually any common approximation, but Ka as being dangerously low so would probably prefer to opt for a full calculation but if Kw is not given surely it is not necessary in an exam/school question?

[Borek]

No intuitive check here, but a quantitative one. [itex][H^+]=\sqrt{C_a K_a}[/itex] can be used if dissociation degree is below 5%.

This is in details described on the page you already know: http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

You are expected to know Kw whether is is given or not. Whether you have to use it or not is another question.

[SomeDude]

you have to remember that a .42M solution of an acid means it is diluted

It doesn't mean anything like that. "Diluted" is a relative term without a solid definition. In some cases 0.42M will be diluted, in some cases it will be quite concentrated, depends on the acid and circumstances.

Sorry about that, what i meant was that it is a weak acid and assumed to be in aqueous solution in this type of problem so you have to find the [H⁺] before the problem can be solved, rather than using 0.42M as the concentration in calculating pH, because it is said to be a weak acid.  it seemed like he might have skipped this step.

Sorry, this was my first post ill make sure im more thorough next time