Which indicator would be the best to use for a titration between 0.20 m hi with 0.10 m ba(oh)2? you will probably need to consul

Question

olga2289 [7]

3 years ago

5

Which indicator would be the best to use for a titration between 0.20 m hi with 0.10 m ba(oh)2? you will probably need to consul

t the appropriate table in the book?

Chemistry

2 answers:

Mars2501 [29] · Answer 1 · 2022-02-12T11:14:27+03:00

1. alizarin yellow R, color change at pH 10-12
2. thymol blue, color change at pH 2-4
3. methyl red, color change at pH 3-5
4. phenolphthalein, color change at pH 8-10
5. bromothymol blue, color change at pH 6-8

These are the list of indicators that we have to choose from them and the correct answer will be 5) Bromothymol blue, due to color change at pH 6-8
Due to both HI is strong acid and Ba(OH)₂ is strong base so the formed salt BaI₂ will has pH = 7 and the equivalence point will be 7 so it will be in the effective range of Bromothymol blue

serious [3.7K] · Answer 2 · 2022-02-12T12:41:04+03:00

The indicator that is used for the endpoint at ${\text{pH}}$ 7 is $\boxed{{\text{bromothymol blue}}}$ .

Further Explanation:

Neutralization reaction:

It is the reaction that occurs between an acid and a base in order to form salt and water. It is named so as it neutralizes the excess amount of hydrogen or hydroxide ions present in the solution. It is used to decrease the acidity in the stomach, wastewater treatment, antacid tablets, and to control the pH of soil. An example of a neutralization reaction is,

${\text{HCl}}+{\text{NaOH}}\to{\text{NaCl}}+{{\text{H}}_2}{\text{O}}$

The indicator is generally a weak acid or weak base that is utilized to determine the ${\text{pH}}$ change of the solution. An indicator when dissolved in water dissociates into the ions and give color. Acid-base indicators are used in neutralization reaction.

The titration between HI and ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ would result in the following neutralization reaction.

$2{\text{HI}}+{\text{Ba}}{\left( {{\text{OH}}}\right)_2}\to{\text{Ba}}{{\text{I}}_2}+2{{\text{H}}_{\text{2}}}{\text{O}}$

According to the balance reaction, 2 moles of HI is needed to completely neutralize 1 mole of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ .

When the volume of solution is 1 L then the number of moles of HI in 0.20 M solution is 0.20 mol. Also, the number of moles of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ present in 0.10 M is 0.10 mol.

Since 1 mole of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ can completely neutralized 2 moles of HI and give 1 mole of ${\text{Ba}}{{\text{I}}_2}$ , therefore, for 0.10 moles of ${\text{Ba}}{\left({{\text{OH}}}\right)_{\text{2}}}$ is completely neutralize 0.20 moles of HI and give 0.10 moles of ${\text{Ba}}{{\text{I}}_2}$ . Since it is neutralization reaction and thus the final ${\text{pH}}$ of the solution is 7.