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Blababa [14]
3 years ago
11

The pH of 0.10 M solution of an acid is 6. What is the percentage ionization of the acid?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
6 0

Hey there!:

HA <=> H⁺ + A⁻

pH = -log[H+] = 6

[ H⁺ ] = 10^-pH

[ H⁺] = 10 ^ -6

[ H⁺ ] = 0.000001 M

Percent dissociation:

[ H⁺ ] / [ HA]o * 100

[ 0.000001 / 0.10 ] * 100

0.00001 * 100 => 0.0010%

Answer  D

Hope that helps!


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Could someone explain how they got this answer, explain step by step plz
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Answer:

6.018 amu

Explanation:

Let 6–Li be isotope A.

Let 7–Li be isotope B.

Let the abundance of 6–Li be A%

Let the abundance of 7–Li be B%

The following data were obtained from the question:

Atomic mass of isotope A (6–Li) =.?

Atomic mass of isotope B (7–Li ) = 7.015 amu.

Abundance of 7–Li (B%) = 92.58%

Abundance of 6–Li (A%) = 100 – B% = 100 – 92.58 = 7.42%

Atomic mass of Lithium = 6.941amu

The atomic mass of isotope A (6–Li) can be obtained as follow:

Atomic mass = [(Mass of A x A%)/100] + [(Mass of B x B%)/100]

6.941 = [(mass of A x 7.42)/100] + [(7.015x92.58)/100]

6.941 = [(mass of A x 7.42)/100] + 6.494487

(mass of A x 7.42)/100 = 6.941 – 6.494487

(mass of A x 7.42)/100 = 0.446513

Mass of A x 7.42 = 100 x 0.446513

Mass of A x 7.42 = 44.6513

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Mass of A = 44.6513 / 7.42

Mass of A = 6.018 amu

Therefore, the mass of 6–Li is 6.018 amu

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