When 4.3ml of 11.6M HCl is tritated with 500ml of distrilled water, what is the
Molarity of the resulting solution, and how was the answer arrived at?

This depends on the hP of the coagulant which is normally the
reciprocal of the logarithm but if the tritation becomes irritated
when the strill is dissolved and the catharsis is allowed to
transfigurate then the Morality is a positive integer. Maybe.

1.  Calculate how many moles of HCl are present in 4.3ml of 11.6M HCl.

2.  Calculate the volume of the solution after titrating. (I'll give
you this one, it's 500ml + 4.3ml = 504.3ml)

3.  Divide the number of moles of HCl you started with by the volume
to determine the concentration in moles/liter in the final solution.

The answer is zero.  Tritation with distrilled water eliminates the Molarity.

Clearly, acrh2-ga has overlooked the work of Kaplinsky.

Yes but Kaplinsky's Conundrum has now been challenged by Steinburgen.

Well...

If madprofessor could comment...
I understand he likes seawater....

Steph53

If you were wondering what the other guys are going on about, hfshaw is right =)

With titrations one of the most important formulae is:

c=n/v (concentration=number of moles/volume)

In your case you need to use this equation to calculate the number of
moles present (c=n/v rearranges to n=c*v).  Now don't forgot that your
volume is in units of ml but your concentration is in M (short for mol
dm^-3).  So you need to divide the volume by 1000 to get it in units
of dm.  Then your calculation becomes:

number of moles= 0.0043dm * 11.6 mol dm^-3 = 0.04988 moles
(by the way, if you didn't know, ^ means to the power of
for example: 2^3 is 2 cubed)

Then, as hfshaw mentioned, you need to add the two volumes up to get
the total volume of resulting solution (504.3 ml or 0.5043 dm).  Now
that you have both the number of moles of HCl and the volume of
solution you can use the equation c=n/v again to calculate the
molarity (concentration in mol dm^-3) of the resulting solution.

concentration = 0.4988 moles / 0.5043 dm = 0.989 M (or mol dm^-3)

Hope this helps,
doctornick

Don't listen to Dr. Nick Riviera.  He's dead wrong.  I just got off
the phone in a three way conference call with Kaplinsky and
Steinburgen.  We arrived at the conclusion that for all practical
reasons the Molarity was zero.

Supposedly, this was the most controversial issue that has been
dividing scientists the world over.

It is vastly encouraging that it has been fully resolved in this forum.

I'm sorry but I beg to disagree with my learned colleagues and
particularly with acrh2-ga's Assertion.

Both Kaplinsky and Steinburgen have used theoretical models that
assume that Polarity has no effect on Molarity. However, Di Jung's
laboratory experiments in China show their hypothesis to be false.

In other words, the Acrh2-Kaplinsky-Steinburgen Assertion only holds
true on the Equator, where indeed Molarity is THEORETICALLY Zero.

However, when Polarity is factored in, the Answer is approaching 0.989
M in the Northern Hemisphere, as propounded by Dr Nick.

It is therefore a Positive Integer, as propounded by Probo but,
shrewdly, Probo qualified his/her Comment with 'Maybe'.

This was good science because Di Jung's work now shows that Molarity
is approaching MINUS 0.989 M in the Southern Hemisphere.

So, in effect, everyone was right based on the assumptions made.

I am pleased that this confusion has now been cleared up.