Question: What is the K_a of HCN if its pK_a=9.31?

Acids differ in their ability to donate H⁺. Stronger acids, such as HCl, react almost completely with water, whereas weaker acids, such as acetic acid (CH₃COOH), react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant (k_a)for the acid-dissociation equilibrium.

Example:

$$\begin{gathered} {\text{HA+H}}_2{\text{O⇄A}}^{-}{\text{H}}_3\text{O} \\ {K}_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]} \end{gathered}$$

Acid strengths normally pk_a uses values rather thank_a values, where the pk_a is the negative common logarithm ofk_a :

pk_a =-log k_a

The stronger acid has a smaller pk_a and a weaker acid has a largerpk_a .

As we know from above step,

pk_a =-log k_a --------eq(1)

Taking antilog on both sides of eq(1),

Antilog (pk_a )= -k_a

Antilog (-9.31) = k_a

10^-9.31 = k_a

4.9 x 10^-10 = k_a

The k_a of HCN is 4.9 x 10^-10 if its pk_a = 9.31.