
What are acid and base dissociation constants?
Answer
511.2k+ views
Hint: We have to know that, acid separation consistent, ${K_a}$ , (otherwise called sharpness steady, or corrosive ionization consistent) is a quantitative proportion of the strength of a corrosive in arrangement. It is the balance consistent for a synthetic response known as separation with regards to corrosive base responses.
Complete answer:
We have to know that a separation steady is a particular sort of harmony consistent with the inclination of a bigger item to isolate reversibly into more modest segments, as when an intricate self-destructs into its segment particles, or when a salt separates into its segment particles. The separation consistent is the reverse of the affiliation steady. In the extraordinary instance of salts, the separation consistent can likewise be called an ionization steady. They are balance constants for the ionization responses of acids and bases.
At the point when a corrosive breaks up in water, it responds as indicated by the condition.
$HA + {H_2}O \rightleftharpoons {H_3}{O^ + } + O{H^ - }$
We can compose a balance steady articulation,
${K_{eq}} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {HA} \right]\left[ {{H_2}O} \right]}}$
Where, the concentration of water is constant, so, that we have to rewrite the above expression,
${K_{eq}} \times \left[ {{H_2}O} \right] = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {HA} \right]}}$
Now we can define a new constant, the acid dissociation constant,
${K_a} = {K_{eq}} \times \left[ {{H_2}O} \right]$
Then, applying the above ${K_a}$ value in the acid dissociation constant formula,
Therefore,
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {HA} \right]}}$
For a base, we can compose the balance response as
$B + {H_2}O \rightleftharpoons B{H^ + } + O{H^ - }$
We can characterize the base separation steady,
${K_b} = {K_{eq}} \times \left[ {{H_2}O} \right]$
So that,
${K_b} = \dfrac{{\left[ {B{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ B \right]}}$
Subsequently, the corrosive and base separation constants are the harmony constants without the term for the grouping of water.
Note:
We have to see that the separation steady is regularly used to depict the liking between a ligand and a protein; for instance how firmly a ligand ties to a specific protein. Ligand-protein affinities are affected by non-covalent intermolecular connections between the two particles.
Complete answer:
We have to know that a separation steady is a particular sort of harmony consistent with the inclination of a bigger item to isolate reversibly into more modest segments, as when an intricate self-destructs into its segment particles, or when a salt separates into its segment particles. The separation consistent is the reverse of the affiliation steady. In the extraordinary instance of salts, the separation consistent can likewise be called an ionization steady. They are balance constants for the ionization responses of acids and bases.
At the point when a corrosive breaks up in water, it responds as indicated by the condition.
$HA + {H_2}O \rightleftharpoons {H_3}{O^ + } + O{H^ - }$
We can compose a balance steady articulation,
${K_{eq}} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {HA} \right]\left[ {{H_2}O} \right]}}$
Where, the concentration of water is constant, so, that we have to rewrite the above expression,
${K_{eq}} \times \left[ {{H_2}O} \right] = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {HA} \right]}}$
Now we can define a new constant, the acid dissociation constant,
${K_a} = {K_{eq}} \times \left[ {{H_2}O} \right]$
Then, applying the above ${K_a}$ value in the acid dissociation constant formula,
Therefore,
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {HA} \right]}}$
For a base, we can compose the balance response as
$B + {H_2}O \rightleftharpoons B{H^ + } + O{H^ - }$
We can characterize the base separation steady,
${K_b} = {K_{eq}} \times \left[ {{H_2}O} \right]$
So that,
${K_b} = \dfrac{{\left[ {B{H^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ B \right]}}$
Subsequently, the corrosive and base separation constants are the harmony constants without the term for the grouping of water.
Note:
We have to see that the separation steady is regularly used to depict the liking between a ligand and a protein; for instance how firmly a ligand ties to a specific protein. Ligand-protein affinities are affected by non-covalent intermolecular connections between the two particles.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

