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Ludmilka [50]
3 years ago
10

Sodium hydrogen carbonate decomposes above 110°c to form sodium carbonate, water, and carbon dioxide. 2nahco3(s) na2co3(s) + h2o

(g) + co2(g) one thousand grams of sodium hydrogen carbonate are added to a reaction vessel, the temperature is increased to 200°c, and the system comes to equilibrium. what happens in this system if another 50 g of sodium carbonate are now added?
Chemistry
1 answer:
Korolek [52]3 years ago
3 0

The given chemical equation is,

2NaHCO_{3}(aq)  Na_{2}CO_{3}(aq)+CO_{2}(g) + H_{2}O(g)

When sodium carbonate is added the system tries to minimize the effect of the additional product added, by shifting the equilibrium such that the extra product is converted back to the reactants. This takes place until the equilibrium is re-established. Therefore, the equilibrium shifts towards the reactants side. The concentrations of sodium carbonate, carbon dioxide and water vapor decrease while the concentration of sodium hydrogen carbonate increases.

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The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

  NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$.

So at the equivalence point, the moles of NaOH added = moles of $HCl$present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$  mole/L

The volume of NaOH added up to the color change = $V_2 \text{  and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$  moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or   when    $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$  mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$, then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$)

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