
How would you complete this Bronsted – Lowry reaction?
\[HC{{O}_{3}}^{-}(aq)+{{H}^{+}}(aq) \to ?\]
Answer
545.4k+ views
Hint:The bronsted acid is defined as the substance which tends to donate the proton to other substances. The bronsted base is defined as the substance which tends to accept the proton from the other substance. The Lowry bronsted concept of acid and bases is applied not only to aqueous solution but also to the non aqueous solution.
Complete step-by-step answer:According to the concept of Bronsted Lowry the acid tends to donate the proton and the base tends to accept the electron from the other substance. When we see that the acid loses its proton then the remaining part of it has the tendency to regain the proton. So it act as a base. Same is the case with base, that is if the base has the tendency to accept the proton then the remaining part of it tends to lose a proton. So it acts as an acid. These pairs of the substance are known as the conjugate acid base pairs.
So the given equation is:
\[HC{{O}_{3}}^{-}(aq)+{{H}^{+}}(aq)? \to \]
Here the \[HC{{O}_{3}}^{-}\]tends to accept the proton so it acts as an Bronsted base.
The\[HC{{O}_{3}}^{-}\] dissociates into the following:
\[HC{{O}_{3}}^{-}C{{O}_{3}}^{2-}+{{H}^{+}}\]
Here we observe that the \[HC{{O}_{3}}^{-}\]loses its proton and acts as the Bronsted acid.
From the equation we get to know that the \[HC{{O}_{3}}^{-}\] can accept the proton to form carbonic acid \[{{H}_{2}}C{{O}_{3}}\]and can also lose one proton to form \[C{{O}_{3}}^{2-}\]. So it can act as the Bronsted acid and Bronsted base also.
So the above equation will be completed when the addition of the hydrogen will be in the \[HC{{O}_{3}}^{-}\]. So the equation will be:
\[HC{{O}_{3}}^{-}(aq)+{{H}^{+}}(aq) \to {{H}_{2}}C{{O}_{3}}\]
So the question mark will be replaced by \[{{H}_{2}}C{{O}_{3}}\].
Note:According to the Lowry Bronsted concept the acid strength depends upon the tendency to lose the protons. The strength of the base depends upon the gain of the protons. The hydrochloric acid is a strong acid that means it has the tendency to donate the protons.
Complete step-by-step answer:According to the concept of Bronsted Lowry the acid tends to donate the proton and the base tends to accept the electron from the other substance. When we see that the acid loses its proton then the remaining part of it has the tendency to regain the proton. So it act as a base. Same is the case with base, that is if the base has the tendency to accept the proton then the remaining part of it tends to lose a proton. So it acts as an acid. These pairs of the substance are known as the conjugate acid base pairs.
So the given equation is:
\[HC{{O}_{3}}^{-}(aq)+{{H}^{+}}(aq)? \to \]
Here the \[HC{{O}_{3}}^{-}\]tends to accept the proton so it acts as an Bronsted base.
The\[HC{{O}_{3}}^{-}\] dissociates into the following:
\[HC{{O}_{3}}^{-}C{{O}_{3}}^{2-}+{{H}^{+}}\]
Here we observe that the \[HC{{O}_{3}}^{-}\]loses its proton and acts as the Bronsted acid.
From the equation we get to know that the \[HC{{O}_{3}}^{-}\] can accept the proton to form carbonic acid \[{{H}_{2}}C{{O}_{3}}\]and can also lose one proton to form \[C{{O}_{3}}^{2-}\]. So it can act as the Bronsted acid and Bronsted base also.
So the above equation will be completed when the addition of the hydrogen will be in the \[HC{{O}_{3}}^{-}\]. So the equation will be:
\[HC{{O}_{3}}^{-}(aq)+{{H}^{+}}(aq) \to {{H}_{2}}C{{O}_{3}}\]
So the question mark will be replaced by \[{{H}_{2}}C{{O}_{3}}\].
Note:According to the Lowry Bronsted concept the acid strength depends upon the tendency to lose the protons. The strength of the base depends upon the gain of the protons. The hydrochloric acid is a strong acid that means it has the tendency to donate the protons.
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