Question

The ${K_a}$ value for the acid $HA$ is $1.0 \times {10^{ - 6}}$ . What is the value of K for the following reaction?
${A^ + } + {H_3}O \leftrightarrow HA + {H_2}O$
A.$1.0 \times {10^{ - 8}}$
B.$1.0 \times {10^8}$
C.$1.0 \times {10^{ - 3}}$
D.$1.0 \times {10^{ - 6}}$

Answer 1

Hint:An enormous ${K_a}$ esteem demonstrates a solid corrosive since it implies the corrosive is to a great extent separated into its particles. An enormous ${K_a}$ esteem likewise implies the development of items in the response is supported. A little ${K_a}$ esteem implies little of the corrosive separates, so you have a powerless corrosive. Powerless acids have a $p{K_a}$ going from 2-1

Complete step by step answer:
Given-
Estimation of ${K_a}$ for the corrosive HA is = $1.0 \times {10^{ - 6}}$
Response of corrosive $HA$ is
$HA + {H_2}O \leftrightarrow {H_3}O + {A^ - }$
${K_a}$ for this response is ${10^{ - 6}}$
Another response is -
${A^ - } + {H_3}{O^ + } \leftrightarrow HA + {H_3}O$
From above responses it is exceptionally certain that
${K^ - } = \dfrac{1}{{{K_a}}}$
Then we get $1 \times {10^-6}$
And hence option D is the correct answer.

Additional information
${K_a}$ , the corrosive ionization consistent, is the harmony steady for substance responses including frail acids in fluid arrangement. The greater the ${K_a}$ value the more it is corrive and the less ${K_a}$ value then it is more vulnerable. For a substance condition of the structure
$HA + {H_2}O \leftrightarrow {H_3}O + {A^ - }(3)$
Where,
$HA$ is the undissociated corrosive and
${A^ - }$ is the form base of the corrosive.

Note:
The amount pH, or "intensity of hydrogen," is a mathematical portrayal of the acidity or basicity of an answer. Arrangements with low pH are the most acidic, and the high pH are generally fundamental.