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I am a chemistry teacher in a Cambridge school. In one of the titrations of sodium carbonate with HCl, there was one question asked:

On heating the sodium carbonate solution before doing the titration, is there any change in the observation/reading?

My opinion is that the rate of reaction will increase due to more collisions among particles due to the temperature change. So the time to completion of the titration should be shortened.

However, the CIE mark scheme suggests that there is no effect, as the concentration remains same.

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    $\begingroup$ For all practical purposes, this reaction has no rate. It just occurs instantly. $\endgroup$ – Ivan Neretin Jan 25 at 9:58
  • $\begingroup$ When in doubt, perform the experiment. Or are you looking for a theoretical explanation of the effect you are asking about? $\endgroup$ – tschoppi Jan 25 at 10:32
  • $\begingroup$ No because bicarbonate isn't involved in the titration. This is what the question is supposed to check $\endgroup$ – Alchimista Jan 25 at 11:53
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I think you might have a different understanding of what constitutes a 'change in the observation/reading' from the textbook, resulting in your confusion.

I am assuming that the textbook is only considering the titration graph per se, pH vs volume of acid added.

You are referring to the time to pH equilibrium during each drop of acid added. For the purposes of all titrations I have encountered, this was never a factor that was discussed. As @Ivan Neretin commented, the reaction is assumed to occur instantly.

As the concentration of the species remains the same, the titration graph will look the same. In other words: The temperature has no influence on the equivalence point.

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With respect to the reaction rate, the rate-limiting step of a titration is typically the mixing of the bulk solution. This is easy to see if you have an indicator dye in the solution. When you add a drop of the acid solution, there will be an immediate local coloring indicative of low pH (also demonstrating, as others said, that the proton transfer is nearly instantaneous). As the solution mixes, the dye returns to colorless (or to whatever the appropriate color for the bulk solution pH). If you aren't actively stirring the solution, this will take much longer than if you have it stirring.

If the "observation/reading" of the experiment is simply the amount of acid required to neutralize the carbonate, then that outcome is unchanged (and presumably that is what the question referred to). However, if the "observation/reading" includes the pKa of the carbonate, that observation will be temperature dependent, although perhaps not detectable within the limited temperature range you are working, depending on the accuracy with which you can measure pH.

Table 9.1 of this paper gives the temperature dependence of both $K_a$ values for carbonate.

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I can not comment because I do not have 50 reputation yet so I post this as an answer even thought it is just an estimated guess. In a titration the reaction should be quite fast so you the result imidiatly. Increasing the temperature would not change much. I think that the Sodium carbonate in Water will form hydrogen carbonate.

$$\ce{Na2CO3_{(aq)} + H2O_{(l)} <=> 2Na+_{(aq)} + OH-_{(aq)} + HCO3-_{(aq)}}$$

When heating $\ce{HCO3-}$ it seems resonable, that CO2 is evaporating forming another $\ce{OH-}$. Therefore you end up with NaOH in the solution which has a different pKs. But I do not know if normal heating is enought for that. The thermal combustion of solid $\ce{Na2CO3}$ to $\ce{Na2O}$ and $\ce{CO2}$ needs around 500°C.

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"My opinion is that the rate of reaction will increase due to more collisions among particles due to the temperature change"

Yes, when the temperature rises, the collision fraction which is equal to or greater than the activation energy ($E_\mathrm{a}$) will be increased.This will increase the rate of the reaction. If the activation energy is very low, close to zero, as for ionic reactions such as precipitation reactions and acid-base neutralization reactions, the fraction of effective collisions does not affect by the temperature rise just with very little value. The **Arrhenius equation **describes the mathematical relationship between the rate constant $(k)$ and temperature $(T)$ as follows: $$k = A \mathrm{ e^{-E_\mathrm a/(R T)}}$$ Where :

$R:$general gases constant

$e:$ The basis of natural logarithm $~\mathrm {e}=2.718$

$A: $Pre-exponential Factor, the value of this term changes slightly with temperature, but can be considered constant.

The other part of the equation $\mathrm{ e^{-E_\mathrm a/(R T)}} $ is a fraction which represents the ratio of effective collisions and increases with higher temperature. The value of this fraction is greater for reactions of low activation energies because a greater proportion of collisions have an energy equivalent to activation energy or superiority, Thus, increasing the proportion of effective collisions, and increasing the value of rate constant$ k$.

The value of this fraction $ \mathrm{ e^{-E_\mathrm a/(R T)}}$ is very small and much less than one for most reactions. When the activation energy ($E_\mathrm{a}$) approaches zero, the value of this fracture approximates the true one as is the case for ionic reactions such as precipitation reactions and acid-base neutralization reactions. In this case, all oriented collisions lead to reactions, so:$\color{red}{\text{ The rate constant value$~ k$ is very high, and the reaction rate is not affected by increasing the temperature .}}$

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