Chapter 18, Problem 97A

Solution

(a)

Interpretation:

The successive ionization equations for polyprotic acid are to be written.

Concept introduction:

According to Arrhenius,

Acid is a substance which gives hydrogen ion in aqueous solution.

Polyprotic acid is the acid which can give more than one proton in aqueous solution.

Acids are classified into monoprotic, diprotic and triprotic acid according to number of hydrogen ions it gives in aqueous solution.

$H_{3}B{O}_3$is polyprotic acid.

Ionization equations are:

$\begin{array}{l}{H}_{3}B{O}_3\to H^{+}+H_{2}B{O}_3{}^{-}\\ H_{2}B{O}_3{}^{-}\to H^{+}+HB{O}_3{}^{2-}\\ HB{O}_3{}^{2-}\to H^{+}+B{O}_3{}^{3-}\end{array}$

(b)

Interpretation:

The successive ionization equations for polyprotic acid are to be written.

Concept introduction:

According to Arrhenius,

Acid is a substance which gives hydrogen ion in aqueous solution.

Polyprotic acid is the acid which can give more than one proton in aqueous solution.

Acids are classified into monoprotic, diprotic and triprotic acid according to number of hydrogen ions it gives in aqueous solution.

$C{H}_{3}COOH$is monoprotic acid.

Ionization equations are:

$C{H}_{3}COOH\to H^{+}+C{H}_{3}CO{O}^{-}$

$C{H}_{3}COOH$is monoprotic acid. It has only one acidic proton. So it dissociates as:

$C{H}_{3}COOH\to H^{+}+C{H}_{3}CO{O}^{-}$

(c)

Interpretation:

The successive ionization equations for polyprotic acid are to be written.

Concept introduction:

According to Arrhenius,

Acid is a substance which gives hydrogen ion in aqueous solution.

Polyprotic acid is the acid which can give more than one proton in aqueous solution.

Acids are classified into monoprotic, diprotic and triprotic acid according to number of hydrogen ions it gives in aqueous solution.

$HN{O}_3$is monoprotic acid.

Ionization equations are:

$HN{O}_3\to H^{+}+N{O}_3{}^{-}$

$HN{O}_3$is monoprotic acid. So, it can give only one hydrogen ion in the solution

(d)

Interpretation:

The successive ionization equations for polyprotic acid are to be written.

Concept introduction:

According to Arrhenius,

Acid is a substance which gives hydrogen ion in aqueous solution.

Polyprotic acid is the acid which can give more than one proton in aqueous solution.

Acids are classified into monoprotic, diprotic and triprotic acid according to number of hydrogen ions it gives in aqueous solution.

$H_{2}Se{O}_3$is polyprotic acid.

Ionization equations are:

$\begin{array}{l}{H}_{2}Se{O}_3\to H^{+}+HSe{O}_3{}^{-}\\ HSe{O}_3{}^{-}\to H^{+}+Se{O}_3{}^{2-}\end{array}$

$H_{2}Se{O}_3$ is diprotic acid. It has two acidic protons which can release in aqueous solution.

So it is protic acid and dissociates as:

$\begin{array}{l}{H}_{2}Se{O}_3\to H^{+}+HSe{O}_3{}^{-}\\ HSe{O}_3{}^{-}\to H^{+}+Se{O}_3{}^{2-}\end{array}$