Question

Why is ${{\text{K}}_{{{\text{a}}_{\text{2}}}}}{\text{ < < }}{{\text{K}}_{{{\text{a}}_{\text{1}}}}}$ for ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ in water? Concentrated ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is added to ${\text{Ca}}{{\text{F}}_{\text{2}}}$?

Answer 1

Hint: The acid dissociation constant is the equilibrium constant of the dissociation reaction of an acid and is denoted by ${{\text{K}}_{\text{a}}}$. This equilibrium constant is a quantitative measure of the strength of an acid in a solution. ${{\text{K}}_{\text{a}}}$ is commonly expressed in units of mol/L.

Step by step answer:
${{\text{K}}_{\text{a}}}$ here denotes acid dissociation constant. Acid dissociation constant is also known as acidity constant or acid-ionization constant. It is a quantitative measure of the strength of an acid in a solution. Acid dissociation constant is the equilibrium constant for any chemical reaction known as dissociation in the context of an acid base reaction. The compound HA is an acid and dissociates into ${{\text{A}}^ - }$ which is the conjugate base of an acid, and hydrogen ion ${{\text{H}}^ + }$.
$\text{HA }\overset{{}}{\leftrightarrows}\text{ }{{\text{A}}^{-}}\text{ + }{{\text{H}}^{\text{+}}}$
This system is said to be in equilibrium when the concentration of its components will not change with the passage of time. This is because, in this reaction both the forward and backward reactions are occurring at the same rate. The dissociation constant ${{\text{K}}_{\text{a}}}$ is defined by the following equation.
${{\text{K}}_{\text{a}}}\text{ = }\dfrac{\left[ {{\text{A}}^{-}} \right]\left[ {{\text{H}}^{\text{+}}} \right]}{\left[ \text{HA} \right]}$
$\Rightarrow \text{ p}{{\text{K}}_{\text{a}}}\text{ = }-\text{lo}{{\text{g}}_{\text{10}}}\text{ }{{\text{K}}_{\text{a}}}\text{ = lo}{{\text{g}}_{\text{10}}}\dfrac{\left[ \text{HA} \right]}{\left[ {{\text{A}}^{-}} \right]\left[ {{\text{H}}^{\text{+}}} \right]}$
In the given question, the equations of dissociation of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ are given as:
${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\overset{{}}{\leftrightarrows}\text{ HS}{{\text{O}}_{\text{4}}}^{-}\text{ + }{{\text{H}}^{\text{+}}}\text{ }{{\text{K}}_{{{\text{a}}_{1}}}}\approx \text{10}$
$\text{HS}{{\text{O}}_{\text{4}}}^{-}\text{ }\overset{{}}{\leftrightarrows}\text{ S}{{\text{O}}_{\text{4}}}^{2-}\text{ + }{{\text{H}}^{\text{+}}}\text{ }{{\text{K}}_{{{\text{a}}_{2}}}}\approx \text{1}\text{.2}\times \text{1}{{\text{0}}^{-2}}$
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is a neutral molecule and hence it readily gives a proton ${{\text{H}}^ + }$, whereas hso4 - is a negatively charged molecule and hence it resists to give a proton. This represents that ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is stronger an acid than ${\text{HS}}{{\text{O}}_{\text{4}}}^ -$. Therefore, in the given question,$\text{ }\!\!~\!\!\text{ }{{\text{K}}_{{{\text{a}}_{1}}}}$, the dissociation constant of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is greater than that of ${\text{HS}}{{\text{O}}_{\text{4}}}^ -$ which is given as$\text{ }\!\!~\!\!\text{ }{{\text{K}}_{{{\text{a}}_{2}}}}$.
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When concentrated sulphuric acid (${\text{conc}}{\text{. }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$) is added to calcium fluoride (${\text{Ca}}{{\text{F}}_{\text{2}}}$) it forms hydrogen fluoride (HF) and calcium sulphate (${\text{CaS}}{{\text{O}}_{\text{4}}}$) . The following reaction takes place:
$\text{Ca}{{\text{F}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\xrightarrow{{}}\text{ CaS}{{\text{O}}_{\text{4}}}\text{ + 2HF}$

Note: Sulphuric acid is a strong mineral acid that is colourless when pure. It is a strong dehydrating agent and from the above solution it is proved that it is a stronger acid than hydrogen sulphate. This nature of sulphuric acid makes it capable of forming weaker acids from salts.