0
$\begingroup$

This question kind of borders on the fields of both semantics and math, as well as chemistry, but I still believe this is the right place to ask.

On a test I took, a question was posed as such:

Via titration we are adding a solution of 25 ml, with 0.5 M NaOH, to 25 ml of HCl. What will the concentration of the acid be?

Now, this question seemed a bit strange to be, and it was definitely new. I'm used to calculating concentrations after dilution and other stoichiometric calculations, but this question seemed like a misworded version of those other questions I've grown familiar to.

So, I asked my teacher if she could clarify. First I asked whether the 25 ml of HCl was pure HCl, or a water solution with HCl in it. She said it was pure HCl. Then I asked if she was looking for the concentration of the acid before ANY reaction between the NaOH and the HCl had occured. To this, she also said yes.

The options were:

0.25 M

0.5 M

0.75 M

1 M

I answered 1 M, my reasoning being that 25 ml of HCl is the same volume as 25 ml of water with NaOH. Obviously, this reasoning is quite flawed, but I didn't have anything better. The answer was wrong. However, when my teacher told me what was meant by the question afterwards, her explanation did not align with the question, nor her previous clarifications. She said that this was a simple case of:

$$C_1 \times V_1 = C_2 \times V_2$$

Where $C_2$ was the unknown, since the concentration of HCl was not specified. I don't see how this makes sense. All the question is saying, is that there is a 25 ml container, with HCl in it. It doesn't even say that this 25 ml consists of a water solution with HCl in it, but a more knowledgeable chemist would've known that HCl cannot exist outside of a water solution (that's what my teacher said at least).

Then the the question says that 25 ml of water, with a concentration of 0.5 M of NaOH is added. And from those two values, I'm supposed to find the concentration of HCl. That basically means:

$0.025$ l of H$_2$O$ \times 0.5$ M NaOH(aq) $= 0.025$ l of water $\times x$ M HCl (aq)

Which would then make the answer 0.5 M. Maybe she wanted me two add the two volumes, which would make sense:

$0.025$ l of H$_2$O$ \times 0.5$ M NaOH(aq) $= 0.05$ l of water $\times x$ M HCl (aq)

Which would then make the answer 0.025, which wasn't even an option.

But even if it was, it still wouldn't make sense to me. Just because you have a solution with a given volume and an unknown concentration, adding another solution with a known volume AND concentration won't just magically let you know what the unknown concentration in the original solution is, now in this new solution. Or am I wrong?

$\endgroup$
5
  • 2
    $\begingroup$ Use rather a meaningful title than a click bait. $\endgroup$ – Poutnik Jan 25 at 17:38
  • $\begingroup$ @Poutnik Is this title better? $\endgroup$ – A. Kvåle Jan 25 at 17:39
  • 4
    $\begingroup$ The answer is either trivial either there is no answer, depending on interpretation of the question. $\endgroup$ – Poutnik Jan 25 at 18:16
  • 1
    $\begingroup$ Your teacher is giving you rather wishy-washy questions and answers. I lean towards the interpretation where 25 ml of 0.5 M NaOH is supposed to exactly neutralise 25 ml of x M HCl, and you are supposed to find x. After all, that's what a titration is about: you add HCl until you neutralise the NaOH exactly, which can be determined using a pH indicator. However, none of this is explicitly stated, and as per Poutnik's comment one could easily say that there is no reasonable answer. $\endgroup$ – orthocresol Jan 25 at 18:16
  • $\begingroup$ The question as it is in the box cannot be answered. And I censor my own opinion about the teacher / teaching. $\endgroup$ – Alchimista Jan 26 at 9:33
4
$\begingroup$

With a lot of assumptions, 0.5 M is a reasonable answer.

Here are the holes you have to fill in.

  1. If it is a titration of NaOH and HCl, it is probably an acid/base titration where we stop the titration when the pH is neutral.

  2. Titrations are often performed to figure out the concentration of a solution. You would titrate a smaller sample of the solution to figure out the concentration. Technically, the concentration of HCl after the titration is zero because all of it is supposed to react as you reach the endpoint of the titration. So it makes sense to ask about the concentration in the original solution.

  3. The stoichiometry of the neutralization of HCl and NaOH is one-to-one: $$\ce{NaOH(aq) + HCl(aq) -> H2O(l) + Na+(aq) + Cl-(aq)}$$

Of course, it is silly not to give any context when posing such a question on a quiz. I won't comment on the statements you attribute to your teacher other than to say that hydrochloric acid always contains water (pure HCl is a gas under ambient conditions).

$\endgroup$
5
  • 1
    $\begingroup$ Sorry, we simultaneously answered it. Hope the OP gets it now. $\endgroup$ – M. Farooq Jan 25 at 18:19
  • 1
    $\begingroup$ @MFarooq No worries. The teacher should read it... $\endgroup$ – Karsten Theis Jan 25 at 18:20
  • $\begingroup$ Good point, it seems that the teacher is teaching the wrong stuff. Nothing annoys me more when I see the dilution formula being used for titrations. $\endgroup$ – M. Farooq Jan 25 at 18:22
  • 1
    $\begingroup$ I think I understand. Since the same amount of NaOH moles is required to neutralize the same amout of HCl molecules, then we know that the amount of moles must be equal in both of the solutions in order for them to neutralize each other. Since the volumes are equal, this means that the molarity of NaOH and HCl will be equal. However, for this to be clear, one must assume that neutralization is the goal. So, if all the values were the same, but HCl was switched out with H$_2$SO$_4$, then the answer would be 0.25, since double the amount of NaOH is needed to neutralize H$_2$SO$_4$? $\endgroup$ – A. Kvåle Jan 25 at 18:33
  • 1
    $\begingroup$ @A.Kvåle, You are right, but always, look at the balanced equation. $\endgroup$ – M. Farooq Jan 25 at 19:28
2
$\begingroup$

The formula $$C_1 \times V_1 = C_2 \times V_2$$ should never ever be used for titrations calculations. It fails miserably in most of the cases (e.g. titration of sulfuric acid with NaOH). This formula only works when the concentrations are in normality, which is considered an obsolete unit in modern chemistry. The proper use of this relation is for calculating dilutions because it is simply a mass balance equation and it works perfectly.

The test question as you wrote is rather poorly worded but basically it is asking you to calculate the concentration of the 25 mL HCl sample if 25 mL of 0.5 M NaOH was required to complete the titration.

All you have to do is to write a balance equation.

From the balanced equation determine the stoichiometric coefficients, i.e., the mole ratio between NaOH and HCl.

Calculate the moles of NaOH = Molarity of NaOH x Volume of NaOH used (L).

Apply the right ratio, and if you have written the balanced equation correctly, you will see the mole ratio is 1:1.

So, m= the moles of HCl = moles of NaOH,

Molarity of HCl = m/Volume of the original HCl sample in L.

You do not care about mixing of volumes until and unless the titration is incomplete, but that would be a separate story.

The correct answer is listed in the choices.

$\endgroup$
2
  • $\begingroup$ But the box question cannot really be answered. I agree it probably meant this, but the "will" makes an already bad question totally non sense.... $\endgroup$ – Alchimista Jan 26 at 9:36
  • $\begingroup$ Either the OP wrote it from memory or he/she translated it into English from another language. $\endgroup$ – M. Farooq Jan 26 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.