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Answer the following questions: (Questions about titration)
a. Why is it important to keep the NaOH solution covered at all times?
b. How will the molarity of NaOH solution be affected by its continued exposure to the atmosphere?
c. The pale pink color of the titration solution at the end point will fade to colorless after several minutes when exposed to the atmosphere. Account for this color change.
d. What volume (in mL) of 0.293 M Ba(OH)2 is required to neutralize 25.00 mL of 0.200M HNO3?

Sagot :

Answer:

Following are the solution to the given question:

Explanation:

For question a:

It is prevented that the atmospheric [tex]CO_2[/tex] through dissolving in the solution and make carbonic acid [tex](H_2CO_3)[/tex] which reacts with the [tex]NaOH:[/tex]

[tex]CO_2+ H_20\to H_2CO_3\\\\H_2CO_3 + NaOH \to NaHCO_3 +H_2O\\\\H_2CO_3 + 2 NaOH \to Na_2CO_3 + 2H_2O\\\\[/tex]

For question b:

For this, the [tex]NaOH[/tex] reacts with the dissolved [tex]CO_2[/tex]  so, the molarity of the [tex]NaOH[/tex] will be decreased.

For question C:

In this, the Phenolphthalein is pink in the basic solution[tex](high \ pH)[/tex] and colorless throughout the acidic solution[tex](low\ pH)[/tex].

if the solution is exposed  from the atmosphere, the [tex]CO_2[/tex] is from the air dissolving in the solution, and making the [tex]H_2CO_3[/tex] that gives the [tex]H^{+}\ ions[/tex]

[tex]\to[/tex] lower pH.

[tex]\to[/tex] colorless phenolphthalein  

For question D:

[tex]Ba(OH)_2 + 2 HNO_3 \to Ba(NO_3)_2+ 2H_2O\\\\[/tex]

Calculating the moles of[tex]HNO_3 = volume \times \text{concentration of} HNO_3\\\\[/tex]

                                                      [tex]= \frac{25}{1000} \times 0.200\\\\= 0.005\ mol\\\\[/tex]

Calculating the moles of  [tex]Ba(OH)_2= \frac{1}{2} \times\text{moles of}\ HNO_3\\\\[/tex]

                                [tex]=\frac{1}{2} \times 0.005\\\\= 0.0025 \ mol\\\\[/tex]

Calculating the volume of [tex]Ba(OH)_2=\frac{moles}{concentration\ of\ Ba(OH)_2}[/tex]

                                   [tex]=\frac{0.0025}{0.0293}\\\\=0.08532\ L\\\\= 85.32 \ mL\\\\= 85.3\ mL[/tex]

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