Question

Compare the relative acidity of 1-molar aqueous solutions of the following acids.

Short answer

${\mathbf{K}}_a$is known as the acid dissociation constant and its value indicates the relative strength of the acid. ${\mathbf{pK}}_a$is defined as the negative logarithm (base $10$) of ${\mathbf{K}}_a$. Mathematically, ${\text{pK}}_{\text{a}}{\text{=-logK}}_{\text{a}}$.value of strong acids is usually zero or even negative while the value of weaker acids are greater than$4$ .

Step-by-step solution

Ka and pKa

${\mathbf{K}}_a$is known as the acid dissociation constant and its value indicates the relative strength of the acid. ${\mathbf{pK}}_a$is defined as the negative logarithm (base $10$) of ${\mathbf{K}}_a$. Mathematically, ${\text{pK}}_{\text{a}}{\text{=-logK}}_{\text{a}}$.value of strong acids is usually zero or even negative while the value of weaker acids are greater than$4$ .

Species in order of decreasing acidity

The strength of an acid or base is given by its ${\text{pK}}_{\text{a}}$value. The smaller localid="1662465510222" ${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$value means stronger is the acid while the larger localid="1662465496640" ${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$value means stronger is the base.

$\begin{array}{l}{\text{pK}}_{\text{a}}{\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}NH}}_{\text{4}}\text{Cl\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}9.26}\\ {\text{pK}}_{\text{a}}\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}HCN\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}9.22}\\ {\text{pK}}_{\text{a}}{\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CH}}_{\text{3}}\text{COOH\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}4.74}\\ {\text{pK}}_{\text{a}}\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}HCOOH\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}3.76}\\ {\text{pK}}_{\text{a}}{\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}-5}\\ {\text{pK}}_{\text{a}}\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}HI\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}-11}\\ {\text{pK}}_{\text{a}}\text{\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}HBr\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}-9}\end{array}$