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I was solving the following problem on a test and I thought the answer was A, but the answer is actually D. Below is the problem and my reasoning:

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Answer choice B is clearly incorrect because the number of drops of indicator largely does not affect the observed end point. Answer choice C is not mathematically valid, so it can also be ruled out.

Between A and D, I thought that A was correct because if some neutralized solution remains in the flask, then the apparent volume of NaOH in the flask is higher and the concentration of NaOH is lower than the actual value. Furthermore, I thought D was incorrect because spilling $\ce{NaOH}$ solution does not change the concentration of the remaining $\ce{NaOH}$.

Can someone please help me understand this question?

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  • $\begingroup$ If your titration reached the end point and you add water, will that change the color of the indicator? If the concentration is all that matters, why even measure the volume in a titration? $\endgroup$ – Karsten Theis Jan 21 at 18:29
  • $\begingroup$ @KarstenTheis What's your point? I'm not getting it $\endgroup$ – DrPepper Jan 21 at 20:21
  • $\begingroup$ The way you know the relation between HCl and NaOH is the stoichiometry of the reaction given by the chemical equation. That means you know how many moles of HCl react with how many moles of NaOH. When you add water (or water and sodium chloride), you are not changing any amounts. In fact, you often add unspecified amounts of water in a titration before you start. Don't get me wrong, there are lots of other scenario where concentration is very important, such as reaction rates, equilibrium reactions, colligative properties, to name just a few. $\endgroup$ – Karsten Theis Jan 21 at 20:53
  • $\begingroup$ @KarstenTheis But if you add water into the titration flask, doesn't that erroneously reduce the concentration of the NaOH in the flask? $\endgroup$ – DrPepper Jan 21 at 21:20
  • $\begingroup$ It does decrease the concentration and increase the volume. You still need to add the same amount of HCl to neutralize it, though. As long as you know the volume of NaOH you put in, and get the amount $n_{OH}$ (how many moles) of hydroxide from the titration, you can calculate the initial concentration. Think about it, when you are titrating, the volume is no longer the original volume of NaOH solution because you add indicator solution before and HCl solution during the titration. $\endgroup$ – Karsten Theis Jan 21 at 21:22
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The answer to the title question is obviously YES.

Regarding the test quiz:

It is simply that titration involves equivalents of substance, not their concentration. The amount of NaOH to be neutralised does not change if you dilute its solution.

All the careful V measurements (volumetric flask and pipetting) that you perform before titration are just to ensure that each trial does indeed contains the same amount of analyte.*

After that amount enters the titration flask, it doesn't matter if you dilute it. Dilution might be even required for practical reason.

*The grid with variable V of NaOH solutions makes little sense. In practice the given sample would have been brought to a standard V and finally divided into trials.

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The concentration of the original NaOH solution is calculated as:

$$c_\ce{NaOH} = c_\ce{HCl} V_\ce{HCl} / V_\ce{NaOH}$$

with $c_\ce{HCl}$ and $V_\ce{HCl}$ the concetration and volume of HCl added during the titration, and $V_\ce{NaOH}$ the volume of NaOH titrated. If any of these values are off, the result is off.

For example, if the HCl has a different concentration than we thought, our result will be wrong. If half of the HCl solution goes on the bench instead of the Erlenmeyer, the result will be wrong. The same goes for spilling some of the NaOH solution after measuring the volume (volume in the Erlenmeyer will be different from the volume claimed to be in it).

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